Sum and Difference of Cubes: Free Response

1. Cube, or not, and which sign goes where . Foundational, 9 points. Question 1 of 5.

A sum or a difference of two perfect cubes always factors the same way: a binomial times a trinomial, with SOAP deciding every sign. This question starts with recognizing the pattern and ends with defending one of its signs.

Part A.
For each expression, say whether it is a sum of two perfect cubes, a difference of two perfect cubes, or neither. For each one that qualifies, give the values of $a$ and $b$ .

$x^3 + 216, \qquad x^3 - 90, \qquad x^2 - 1000$

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part B.
Factor $64x^3 + 27$ completely.

Write the expression An equation or an expression is enough here. Show how you built it. 3 points
Part C.
In your factorization from part B, explain why the trinomial's middle term must carry the sign SOAP predicts, arguing from the cancellation that proves the cube identity rather than by citing the rule by name. Then say specifically what would go wrong in the expansion if that sign were reversed.

Carry your own answer forward Use whichever trinomial you found in part B; the cancellation argument works the same way regardless of the exact coefficients.

Explain why it works A sentence or two. Reasons, not steps. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
Two separate questions decide whether an expression matches the pattern: is each term individually a perfect cube, and are the two terms joined by a plus or a minus? An expression can fail on either test alone.
Hint 2 of 4 · Part A
Check $90$ and $1000$ against the list of small perfect cubes ( $1,8,27,64,125,216,343,512,729,1000$ ) before deciding whether either expression qualifies.
Hint 3 of 4 · Part B
Cube-root the coefficient and the variable separately: $64x^3$ needs a whole number whose cube is $64$ , paired with the variable $x$ itself.
Hint 4 of 4 · Part C
Multiply part B's factorization out term by term and watch which pairs of terms are exact opposites of each other. Then redo that same multiplication with the middle sign flipped and see what survives instead.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Test each expression against two requirements: both terms must be perfect cubes, and they must be joined by a plus or a minus sign.

For $x^3+216$ , $x^3=(x)^3$ and $216=6^3$ , and the terms are added.

$x^3+216=x^3+6^3$

Both conditions hold, so this is a sum of cubes with $a=x$ and $b=6$ .

For $x^3-90$ , the first term $x^3$ is a cube, but $90$ is not: $4^3=64$ and $5^3=125$ , and $90$ falls strictly between them. No whole number cubes to $90$ , so this expression does not match the pattern.

For $x^2-1000$ , the number $1000=10^3$ is a perfect cube, but $x^2$ is a square, not a cube. A cube pattern needs BOTH terms to be cubes, so a mismatched power breaks it even when one term qualifies. This one is a difference of squares in disguise, not a difference of cubes.

Part B

Take the cube root of each term. Since $64x^3=(4x)^3$ and $27=3^3$ , this is a sum of cubes with $a=4x$ and $b=3$ . By SOAP, the binomial keeps the plus sign, and the trinomial's middle term takes the opposite (minus) sign, with a positive last term.

$64x^3+27=(4x+3)\bigl((4x)^2-(4x)(3)+3^2\bigr)=(4x+3)(16x^2-12x+9).$

The first trinomial term is $(4x)^2=16x^2$ , the middle term is a single $ab=12x$ with a minus sign, and the last term is $3^2=9$ .

Part C

The SOAP rule is not an arbitrary mnemonic; it reports what has to happen for the expansion to collapse to just two terms. Expand $(4x+3)(16x^2-12x+9)$ and watch the middle terms:

$(4x+3)(16x^2-12x+9)=64x^3-48x^2+36x+48x^2-36x+27.$

The $-48x^2$ and $+48x^2$ are exact opposites and cancel, and so do $+36x$ and $-36x$ , leaving $64x^3+27$ .

If the trinomial's middle sign were flipped to plus instead, the product $(4x+3)(16x^2+12x+9)$ would pair $+48x^2$ with $+48x^2$ (reinforcing, not cancelling) and $+36x$ with $+36x$ . Neither middle pair would vanish, so the product would carry leftover $x^2$ and $x$ terms and could not equal the two-term expression $64x^3+27$ . The minus sign is forced by the cancellation, not chosen by convention.

In one line

$x^3+216$ is a sum of cubes ( $a=x,b=6$ ); $x^3-90$ and $x^2-1000$ are neither. $64x^3+27=(4x+3)(16x^2-12x+9)$ , and the trinomial's middle sign must be minus because only that sign makes the expansion's cross terms cancel in opposite pairs.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Correctly classifies all three expressions as a sum of cubes, a difference of cubes, or neither. . Worth 2 points.

For the expression that qualifies, states the correct values of $a$ and $b$ , and for at least one rejected expression states which of the two requirements (both terms cubes, joined by plus or minus) fails. . Worth 1 point.

Part B 3 points

Correctly identifies the cube root of each term and keeps the binomial's sign the same as the original expression. . Worth 2 points.

Forms the trinomial with the correct first term, a single $ab$ middle term (not doubled), and the correct last term, with the correct sign on the middle term. . Worth 1 point.

Part C 3 points

Explains the sign requirement by appealing to what the expansion's cross terms must do (cancel to opposites), not merely by citing SOAP as a rule to follow. . Worth 2 points. needs an explanation, not just an answer

Names the specific pair of cross terms that fails to cancel if the middle sign were flipped. . Worth 1 point.

2. A shortcut that looks right and is not . Reasoning, 9 points. Question 2 of 5.

Somebody claims: for all real numbers $a$ and $b$ , $a^3+b^3=(a+b)^3$ , so you can skip the sum-of-cubes pattern entirely and just cube the sum. This question asks you to test that claim and say exactly what your test does and does not prove.

Part A.
Disprove the claim $a^3+b^3=(a+b)^3$ for all real numbers $a$ and $b$ : choose one specific pair of numbers, evaluate both sides on that pair, and show the two values differ.

Construct a counterexample Give one specific case, and show it breaks the claim. 4 points
Part B.
Using the same pair from part A, verify that the correct identity, $a^3+b^3=(a+b)(a^2-ab+b^2)$ , does give the right value of $a^3+b^3$ .

Carry your own answer forward Use whichever pair you chose in part A. The check works the same way for any real pair, so this part is not lost if part A's pair was different from the one shown here.

Solve and show your work Write each step out, and end with the value and its units. 2 points
Part C.
State precisely what your counterexample in part A does and does not establish about the claim $a^3+b^3=(a+b)^3$ . Then name the extra terms that $(a+b)^3$ carries beyond $a^3+b^3$ .

Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 3 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
This claim is about every pair of real numbers, so a single pair on which the two sides give different values is all it takes to break it. Try small numbers and compute both sides honestly before assuming anything about them.
Hint 2 of 3 · Part A
Pick two small numbers that are nonzero and not opposites of each other, cube each one and add, and separately add them first and cube the sum. The order the claim asserts does not matter is exactly what you are testing.
Hint 3 of 3 · Part C
Expand $(a+b)^3$ the way you would expand any binomial raised to the third power, by multiplying $(a+b)(a+b)(a+b)$ out fully, and compare term by term with $a^3+b^3$ .

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

A claim about ALL real numbers falls to a single pair on which it fails, so pick a small pair and evaluate both sides honestly.

Take $a=2$ and $b=3$ . The left side is

$a^3+b^3=2^3+3^3=8+27=35.$

The right side cubes the sum first:

$(a+b)^3=(2+3)^3=5^3=125.$

Since $35\neq125$ , this pair refutes the claim. Any pair with $a$ , $b$ , and $a+b$ all nonzero does the same job; the point is that a single honest evaluation is enough to kill a claim about ALL real numbers.

Part B

Substitute the same pair into the right side of the correct identity and simplify.

$(a+b)(a^2-ab+b^2)=(2+3)\bigl(2^2-(2)(3)+3^2\bigr)=5(4-6+9)=5(7)=35.$

This matches $a^3+b^3=35$ from part A, so unlike the cubed-sum shortcut, the SOAP-built identity survives the check.

Part C

A single counterexample is enough to destroy a FOR ALL claim, but it settles nothing beyond that. It does not show the two expressions never agree; in fact $a=0$ (or $b=0$ ) makes both sides equal, since then $a^3+b^3=b^3=(a+b)^3$ trivially. What the counterexample rules out is only the universal version of the claim.

Expanding the right side shows exactly what the shortcut misses:

$(a+b)^3=a^3+3a^2b+3ab^2+b^3.$

Compared with $a^3+b^3$ , the cube of a sum carries two extra middle terms, $3a^2b$ and $3ab^2$ , that the true factoring identity never introduces. Those are exactly what made $125$ and $35$ different in part A.

In one line

The pair $a=2,b=3$ gives $a^3+b^3=35$ but $(a+b)^3=125$ , so the claim fails; the correct identity $(a+b)(a^2-ab+b^2)$ does give $35$ . The counterexample refutes only the universal claim (the two sides agree when $a=0$ , $b=0$ , or $a=-b$ ), and $(a+b)^3$ carries two extra terms, $3a^2b+3ab^2$ , that $a^3+b^3$ does not.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Chooses a specific numerical pair, rather than describing in general terms when the claim might fail. . Worth 2 points.

Evaluates both $a^3+b^3$ and $(a+b)^3$ correctly on that pair and reaches two different values. . Worth 2 points.

Part B 2 points

Substitutes the SAME pair from part A into the correct identity, rather than a new pair. . Worth 1 point.

Reports that the two sides agree, and connects that agreement back to the disagreement found in part A. . Worth 1 point.

Part C 3 points

States the correct logical scope: a single counterexample refutes the universal claim, without asserting the stronger claim that the two sides always disagree. . Worth 2 points.

Names the two extra cross terms that $(a+b)^3$ carries beyond $a^3+b^3$ , rather than describing the difference only in general terms. . Worth 1 point.

3. Uncover the cubes, then finish the job . Application, 11 points. Question 3 of 5.

An expression is not a sum or difference of cubes until you check for a common factor first: pulling one out can reveal the pattern hiding underneath. Once you have it, the zero-product property and the integer-pair test can finish the rest.

Part A.
Factor $2x^3 - 1458$ completely.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part B.
Solve $2x^3 - 1458 = 0$ for real $x$ , using your factorization from part A. You may use, without proving it, that the quadratic factor is positive for every real $x$ .

Carry your own answer forward Solve using whichever factorization you found in part A. The method (zero-product property, then the given fact about the quadratic factor) is the same regardless of the exact numbers.

Solve and show your work Write each step out, and end with the value and its units. 3 points
Part C.
Consider the trinomial $x^2 + 6x + 36$ , the trinomial factor that arises when $x^3 - 216$ is factored as a difference of cubes. Show that $x^2 + 6x + 36$ has no factorization into two binomials with integer coefficients.

Justify your claim State the claim, then give the reason it has to be true. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
Before you look for a cube pattern anywhere in this question, check every expression for a common factor first. Pulling one out is often what makes a hidden pattern visible.
Hint 2 of 4 · Part A
Both terms of $2x^3-1458$ share a factor of $2$ . Divide it out and look at what remains before deciding whether a cube identity applies.
Hint 3 of 4 · Part B
A product of factors is zero exactly when at least one factor is zero. Check each factor from part A's result in turn, using the given fact to dispose of one of them quickly.
Hint 4 of 4 · Part C
List every pair of positive integers that multiplies to $36$ , and check each pair's sum against the target. Do not stop after checking just one or two pairs.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Neither $2x^3$ nor $1458$ is itself a perfect cube, so check for a common factor first. Both terms are even, and dividing by $2$ leaves

$2x^3-1458=2(x^3-729).$

Now $x^3-729=x^3-9^3$ is a difference of cubes with $a=x$ and $b=9$ . By SOAP, the binomial keeps the minus sign, the middle term takes the opposite (plus) sign, and the last term is positive:

$x^3-729=(x-9)(x^2+9x+81).$

Keeping the $2$ out front, the complete factorization is $2(x-9)(x^2+9x+81)$ .

Part B

With the equation in factored form, the zero-product property applies to each factor in turn.

$2(x-9)(x^2+9x+81)=0$

The constant factor $2$ is never $0$ . Setting the next factor to $0$ gives $x-9=0$ , so $x=9$ . The quadratic factor is positive for every real $x$ (given), so it contributes no further real solution. The only real solution is $x=9$ .

Part C

To factor $x^2+6x+36$ as $(x+p)(x+q)$ with integers $p,q$ you would need $pq=36$ and $p+q=6$ . List the positive integer pairs that multiply to $36$ and check their sums.

$1\times36,\ 2\times18,\ 3\times12,\ 4\times9,\ 6\times6$

Their sums are $37,20,15,13,12$ . None of these is $6$ . A pair of negative integers multiplying to $36$ would give a positive product but a negative sum (at best $-12$ ), which also misses $6$ . So no integer pair satisfies both conditions at once, and $x^2+6x+36$ does not factor into two binomials with integer coefficients.

In one line

$2x^3-1458=2(x-9)(x^2+9x+81)$ , and the only real solution of $2x^3-1458=0$ is $x=9$ , since the quadratic factor is always positive. The trinomial $x^2+6x+36$ has no integer binomial factorization, because no pair of integers multiplying to $36$ sums to $6$ .

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Pulls out the common factor of $2$ BEFORE checking for a cube pattern, rather than trying to apply the cube identity to the original expression. . Worth 2 points.

Produces the complete factorization with the correct signs on both the binomial and the trinomial's middle term. . Worth 2 points.

Part B 3 points

Applies the zero-product property to the factored equation and solves the linear factor for $x$ . . Worth 2 points.

Uses the given fact about the quadratic factor to correctly conclude it contributes no additional real solution, rather than leaving it unaddressed. . Worth 1 point.

Part C 4 points

Sets up the correct test: an integer pair must multiply to $36$ AND sum to $6$ , not merely satisfy one of the two conditions. . Worth 1 point.

Checks the integer pairs multiplying to $36$ against the required sum and shows systematically that none works, including ruling out negative pairs. . Worth 2 points. needs an explanation, not just an answer

States the conclusion that the trinomial has no integer binomial factorization. . Worth 1 point.

4. Repairing a proposed cube factorization . Application, 11 points. Question 4 of 5.

A proposed factorization is not automatically right just because the binomial looks correct. This question asks you to find the one change that fixes a wrong trinomial, apply the pattern correctly elsewhere, and explain what the wrong trinomial actually was.

Part A.
A proposed factorization of $8x^3 - 343$ is $(2x - 7)(4x^2 + 28x + 49)$ . This is not correct. Identify the single change needed to fix the trinomial, and write the correct factorization.

Find and correct the error Say which line first goes wrong, why it is wrong, and then do it correctly. 3 points
Part B.
Factor $125x^3 + 8$ completely.

Write the expression An equation or an expression is enough here. Show how you built it. 4 points
Part C.
The wrong trinomial from part A is actually a correct factorization of something else. Identify what kind of expression it correctly factors, write that factorization, and explain why doubling the middle term produces exactly that pattern instead of the cube trinomial.

Explain why it works A sentence or two. Reasons, not steps. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 3
Two different patterns in this chapter both produce a trinomial with an $ab$ -type middle term, and only one of them doubles it. Deciding which pattern applies here is the whole question.
Hint 2 of 3 · Part A
Check the binomial factor first (is its sign right for a difference?), and only then check the trinomial's three terms one at a time against what the cube pattern requires.
Hint 3 of 3 · Part C
Test whether the wrong trinomial fits the perfect-square pattern $a^2+2ab+b^2$ : are its first and last terms perfect squares, and is the middle term exactly twice the product of their roots?

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

Check the binomial first: $8x^3=(2x)^3$ and $343=7^3$ , so $a=2x$ and $b=7$ , and the binomial $(2x-7)$ correctly keeps the minus sign of the original difference. That part is right.

The trinomial is where the error sits. A cube trinomial's middle term is always a SINGLE $ab$ , never the doubled $2ab$ that belongs to a perfect-square trinomial. Here $ab=(2x)(7)=14x$ , so the correct middle term is $14x$ , not $28x=2ab$ . By SOAP the sign is opposite the binomial's minus, so it is plus.

$8x^3-343=(2x-7)(4x^2+14x+49)$

The proposed trinomial doubled the middle term, mistaking the cube pattern for the perfect-square pattern.

Part B

Take the cube root of each term. Since $125x^3=(5x)^3$ and $8=2^3$ , this is a sum of cubes with $a=5x$ and $b=2$ . By SOAP, the binomial keeps the plus sign, and the trinomial's middle term is a single $ab$ with the opposite (minus) sign:

$125x^3+8=(5x+2)\bigl((5x)^2-(5x)(2)+2^2\bigr)=(5x+2)(25x^2-10x+4).$

The first trinomial term is $(5x)^2=25x^2$ , the middle term is $ab=10x$ (not $20x$ ), and the last term is $2^2=4$ .

Part C

Test whether $4x^2+28x+49$ is a perfect-square trinomial: its first and last terms are $(2x)^2=4x^2$ and $7^2=49$ , both perfect squares, and twice their product is $2(2x)(7)=28x$ , matching the middle term exactly.

$4x^2+28x+49=(2x+7)^2$

So the wrong trinomial is not nonsense: it is the genuinely correct expansion of $(2x+7)^2$ , just not the pattern this problem calls for. The perfect-square identity $(a+b)^2=a^2+2ab+b^2$ doubles its middle term because the SAME cross-product $ab$ appears twice when $(a+b)(a+b)$ is expanded, once from each order of multiplying the two terms. The cube trinomial's middle term never doubles that way, because it comes from a DIFFERENT expansion, a binomial times a DIFFERENT trinomial, in which only one cross term of that size survives after the rest cancel. Reaching for the familiar doubled middle term from squaring a binomial, instead of rederiving the cube trinomial's own single $ab$ , is exactly the slip in part A.

In one line

The proposed trinomial doubled the middle term; the correct factorization is $8x^3-343=(2x-7)(4x^2+14x+49)$ . Also, $125x^3+8=(5x+2)(25x^2-10x+4)$ . The wrong trinomial from part A is actually $(2x+7)^2$ , a perfect square, because doubling the middle term is exactly the perfect-square rule, not the cube rule.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Identifies the specific numerical error in the trinomial's middle term, naming which coefficient is wrong and why, rather than a vague statement that the trinomial is wrong or a misdiagnosis of the binomial. . Worth 2 points.

Writes the fully corrected factorization with the right coefficient on the middle term. . Worth 1 point.

Part B 4 points

Correctly identifies the cube root of each term and keeps the binomial's plus sign. . Worth 2 points.

Forms the trinomial with a single $ab$ (not doubled) and the correct sign on the middle term. . Worth 2 points.

Part C 4 points

Verifies that the wrong trinomial's three terms match the pattern of squaring a two-term binomial (a perfect square first term, a perfect square last term, and a middle term equal to twice the product of their roots), rather than the cube pattern's single $ab$ . . Worth 2 points.

Explains why the perfect-square pattern doubles its middle term while the cube trinomial does not, connecting the two different expansions rather than just asserting the rule. . Worth 2 points. needs an explanation, not just an answer

5. Deriving the difference identity, and what squares cannot do . Reasoning, 10 points. Question 5 of 5.

You have used the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$ throughout this lesson. This question asks you to establish it yourself the way you would check any proposed factorization, by expanding and watching what survives, and then to say what a sum of squares would need in order to do the same trick.

Part A.
Prove that $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ for every real number $a$ and $b$ , by expanding the right side and showing exactly which terms cancel.

Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 4 points
Part B.
Use the identity to factor $x^3 - 512$ completely.

Write the expression An equation or an expression is enough here. Show how you built it. 2 points
Part C.
A sum of squares, $a^2+b^2$ , does not factor into two real linear factors (a fact from the difference-of-squares lesson), while part A shows that a difference of cubes does factor, and the lesson shows a sum of cubes does too. Test the most natural square candidate anyway: expand $(a+b)^2$ and $(a-b)^2$ , and say exactly why neither equals $a^2+b^2$ except in a degenerate case. Then say, structurally, why the cube identity from part A never runs into that same obstacle.

Compare the two methods Say what each one costs you, and when you would reach for it. 4 points

Hints

One at a time, each one a step further than the last. Take only as many as you need.

Hint 1 of 4
A proof for ALL real numbers cannot rely on any specific pair; the whole argument has to run in letters, watching which terms are exact opposites of each other.
Hint 2 of 4 · Part A
Multiply every term of $a-b$ against every term of $a^2+ab+b^2$ separately before combining anything. You should end up with six terms before any cancelling happens.
Hint 3 of 4 · Part B
Find the whole number whose cube is $512$ before applying the identity; it is smaller than you might guess.
Hint 4 of 4 · Part C
Expand $(a+b)^2$ and $(a-b)^2$ the ordinary way and compare each to $a^2+b^2$ term by term. Then look back at your part A expansion and count how many separate cross terms of size $ab$ appeared there, compared to how many appear when you square a binomial.

Answer and solution

Check your answer first. If it is wrong, go back to your paper: the worked solution will still be here.

Worked solution

Part A

The claim is about every real $a$ and $b$ , so the argument has to be a general expansion, not a numerical check. Multiply each term of $a-b$ against each term of $a^2+ab+b^2$ :

$(a-b)(a^2+ab+b^2)=a^3+a^2b+ab^2-a^2b-ab^2-b^3.$

Group the four middle terms into matched pairs: $+a^2b$ and $-a^2b$ are exact opposites and cancel, and $+ab^2$ and $-ab^2$ are exact opposites and cancel as well.

$a^3+a^2b+ab^2-a^2b-ab^2-b^3=a^3-b^3.$

Only the two cubes survive, for every choice of $a$ and $b$ , since nothing about the cancellation depended on their particular values. So $(a-b)(a^2+ab+b^2)=a^3-b^3$ always, which read right to left is the factoring rule for a difference of cubes.

Part B

Since $512=8^3$ , this is a difference of cubes with $a=x$ and $b=8$ . Substitute directly into the identity from part A:

$x^3-512=x^3-8^3=(x-8)(x^2+8x+64).$

The trinomial's first term is $x^2$ , the middle term is a single $ab=8x$ with the opposite (plus) sign, and the last term is $8^2=64$ .

Part C

Expand both perfect-square candidates:

$(a+b)^2=a^2+2ab+b^2,\qquad(a-b)^2=a^2-2ab+b^2.$

Each carries an extra $\pm2ab$ term beyond $a^2+b^2$ . For either expression to equal $a^2+b^2$ , that extra term would have to vanish, which happens only when $ab=0$ , that is, when $a=0$ or $b=0$ . For a general pair with both $a$ and $b$ nonzero, neither perfect square equals $a^2+b^2$ , so this obvious route to a sum-of-squares factorization fails.

The cube case is structurally different. In part A, the trinomial's middle term is a single $ab$ , and when the binomial is multiplied through, that single $ab$ meets its own opposite from the OTHER cross term of the expansion ( $+a^2b$ against $-a^2b$ , and separately $+ab^2$ against $-ab^2$ ) and they cancel EXACTLY, for every $a$ and $b$ , with nothing left needing to be zero. The square candidate produces only ONE $ab$ -type term with nothing to pair against, so it survives instead of cancelling. That is the structural difference: the cube identity's middle terms are built to cancel each other, while the square candidate's middle term has no partner and simply stays.

In one line

For all real $a,b$ , $(a-b)(a^2+ab+b^2)=a^3-b^3$ , since the four middle terms cancel in matched pairs. Applying it, $x^3-512=(x-8)(x^2+8x+64)$ . And $(a\pm b)^2=a^2\pm2ab+b^2$ equals $a^2+b^2$ only when $a=0$ or $b=0$ , because the square candidate's middle term has no second cross term to cancel against, unlike the cube trinomial's.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Expands the full product with every term shown, then groups the four middle terms into matched pairs and shows both pairs are exact opposites, for arbitrary $a$ and $b$ rather than specific numbers. . Worth 3 points. needs an explanation, not just an answer

States the concluding factoring rule the cancellation establishes. . Worth 1 point.

Part B 2 points

Recognizes $512$ as a perfect cube and correctly identifies the cube root that pairs with it. . Worth 1 point.

Applies the identity correctly, with a single $ab$ and the opposite sign on the middle term. . Worth 1 point.

Part C 4 points

Expands both $(a+b)^2$ and $(a-b)^2$ correctly and identifies the exact condition under which either equals $a^2+b^2$ , not just a vague statement that they sometimes agree. . Worth 2 points.

Explains the structural difference by counting how many cross terms of matching size each expansion produces and whether each one has a partner to cancel against, rather than simply restating that squares do not factor. . Worth 2 points. needs an explanation, not just an answer

1. Cube, or not, and which sign goes where . Foundational, 9 points. Question 1 of 5.

2. A shortcut that looks right and is not . Reasoning, 9 points. Question 2 of 5.

3. Uncover the cubes, then finish the job . Application, 11 points. Question 3 of 5.

4. Repairing a proposed cube factorization . Application, 11 points. Question 4 of 5.

5. Deriving the difference identity, and what squares cannot do . Reasoning, 10 points. Question 5 of 5.

Marking rubrics

1. Cube, or not, and which sign goes where 9 points

2. A shortcut that looks right and is not 9 points

3. Uncover the cubes, then finish the job 11 points

4. Repairing a proposed cube factorization 11 points

5. Deriving the difference identity, and what squares cannot do 10 points