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Sum and Difference of Cubes: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Finding the missing constant

The expression $8x^3+C$ has a sum-of-cubes factorization containing the binomial $2x+5$ . Find $C$ .
1. Hint 1
  A sum-of-cubes factorization has the form $(a+b)(a^2-ab+b^2)$ , where $a$ and $b$ are the cube roots of the two terms.
2. Hint 2
  Match $2x+5$ against $a+b$ to read off $a$ and $b$ , then write $C$ in terms of $b$ .
Answer

$C=125$ .

Full solution

The cube roots are $2x$ and $5$ .
The constant in the original sum is therefore
$C=5^3,$
giving $C=125$ .
The complete product is $(2x+5)(4x^2-10x+25)$ , whose constant is indeed $125$ .

Answer

$C=125$ .

Key idea

The second term of the binomial is the cube root of the constant term, so cubing it recovers that constant.
Problem 1: Mark complete
Problem 2 An adjusted product

Simplify $(6x-1)(36x^2+6x+1)+1$ .
1. Hint 1
  The product has the sign pattern for a difference of cubes.
2. Hint 2
  Identify the two cube roots before combining the added constant.
Answer

$216x^3$ .

Full solution

The factors match the difference of cubes with terms $6x$ and $1$ , so the product is
$216x^3-1.$
Adding $1$ cancels the constant, leaving $216x^3$ .
At $x=1$ the original expression is $5\cdot43+1=216$ , which matches $216(1)^3$ .

Answer

$216x^3$ .

Key idea

Recognizing a cube identity can expose a constant cancellation without a long expansion.
Problem 2: Mark complete
Problem 3 A complete factorization

Factor $81x^3-375$ completely with integer coefficients.
1. Hint 1
  Neither term is a perfect cube as written, so check for a common factor first.
2. Hint 2
  Once the common factor is out, what is left inside the parentheses is a difference of cubes, and SOAP places its three signs.
Answer

$3(3x-5)(9x^2+15x+25)$ .

Full solution

Both terms are divisible by $3$ , since their digit sums $9$ and $15$ are.
Taking $3$ out leaves
$27x^3-125$
inside the parentheses, and those two terms share no further common factor, so $3$ was the greatest common factor.
Since $27x^3=(3x)^3$ and $125=5^3$ , that is a difference of cubes with $a=3x$ and $b=5$ .
SOAP makes the binomial $3x-5$ , gives the trinomial the opposite middle sign with a single $ab=15x$ , and keeps $b^2=25$ positive, so the complete factorization is
$3(3x-5)(9x^2+15x+25).$
Once a common factor is out and the cube identity is applied, the factorization is finished, so the trinomial is left alone.
At $x=1$ the original expression is $-294$ , and the product gives
$3(-2)(49)=-294.$

Answer

$3(3x-5)(9x^2+15x+25)$ .

Key idea

SOAP places all three signs: the binomial takes the same sign, the middle term the opposite sign, and the last term stays positive.
Problem 3: Mark complete
Problem 4 Combine, then factor

Subtract $3x^3+1536$ from $5x^3+2560$ , write the result as a completely factored product with integer coefficients, and evaluate it at $x=1$ .
1. Hint 1
  Combine like terms in the difference first.
2. Hint 2
  Remove the numerical common factor, then recognize the remaining sum of cubes.
Answer

$2(x+8)(x^2-8x+64)$ ; value at $x=1$ : $1026$ .

Full solution

Subtracting term by term gives
$2x^3+1024.$
Removing the greatest common factor $2$ leaves $x^3+512$ inside the parentheses.
Since $512=8^3$ , that is a sum of cubes with $a=x$ and $b=8$ , so the complete product is
$2(x+8)(x^2-8x+64).$
At $x=1$ the factors give
$2(9)(57)=1026.$
Subtracting the two original expressions at $x=1$ gives $2565-1539=1026$ , which checks the result.

Answer

$2(x+8)(x^2-8x+64)$ ; value at $x=1$ : $1026$ .

Key idea

Combine like terms before factoring: a difference of two expressions can show a cube pattern that neither one shows on its own.
Problem 4: Mark complete
Problem 5 Two related cube products

Let $P=(x+7)(x^2-7x+49)$ and $Q=(x-7)(x^2+7x+49)$ . Find $P-Q$ and $P+Q$ , each with like terms combined.
1. Hint 1
  One product is a sum of cubes and the other is a difference of cubes.
2. Hint 2
  Rewrite each using its two cube terms, then add and subtract.
Answer

$P-Q=686$ ; $P+Q=2x^3$ .

Full solution

The cube identities give
$P=x^3+343$
and
$Q=x^3-343.$
Subtracting removes the cube terms and leaves $686$ , while adding removes the constants and leaves $2x^3$ .
At $x=1$ the values are $P=344$ and $Q=-342$ , so $P-Q=686$ and $P+Q=2$ , which is $2(1)^3$ .

Answer

$P-Q=686$ ; $P+Q=2x^3$ .

Key idea

Adding or subtracting a matched pair of cube products leaves twice the variable cube or twice the constant cube.
Problem 5: Mark complete
Problem 6 A packing box

A rectangular box has dimensions $x+1$ cm, $x^2-x+1$ cm, and $3$ cm, where $x>0$ . Its volume is $84$ cubic cm. Find $x$ , showing how the product of the first two dimensions simplifies.
1. Hint 1
  The first two dimensions form the factors of a sum of cubes.
2. Hint 2
  Multiply that two-term result by the remaining dimension, then isolate the cube of $x$ .
Answer

$x=3$ ; $(x+1)(x^2-x+1)=x^3+1$ .

Full solution

The product of the first two dimensions is $x^3+1$ .
The volume condition is
$3(x^3+1)=84.$
Dividing by $3$ and subtracting $1$ gives $x^3=27$ , so $x=3$ .
The dimensions are then $4$ cm, $7$ cm, and $3$ cm, whose product is $84$ cubic cm.

Answer

$x=3$ ; $(x+1)(x^2-x+1)=x^3+1$ .

Key idea

A cube factorization can simplify a three-dimensional volume condition before solving for a parameter.
Problem 6: Mark complete
Problem 7 A missing multiplier

Write $6x^4-162x$ as $2x$ times a second expression, then factor that second expression completely with integer coefficients.
1. Hint 1
  Take the greatest common factor out of both terms before splitting off the $2x$ .
2. Hint 2
  After $6x$ comes out, what is left inside the parentheses is a difference of cubes.
Answer

The second expression is $3x^3-81$ , that is $3(x^3-27)$ , which factors as $3(x-3)(x^2+3x+9)$ .

Full solution

The two terms share the factor $6x$ , and taking it out leaves
$x^3-27$
inside the parentheses.
Writing $6x$ as $2x$ times $3$ shows that the second expression is $3x^3-81$ , which is $3(x^3-27)$ .
Since $27=3^3$ , that expression is a difference of cubes with $a=x$ and $b=3$ , so the complete factorization is
$3(x-3)(x^2+3x+9).$
No two integers have product $9$ and sum $3$ , so the trinomial does not factor further.
Multiplying by $2x$ returns $6x^4-162x$ , so nothing was divided away.
At $x=2$ both forms give $-228$ .

Answer

The second expression is $3x^3-81$ , that is $3(x^3-27)$ , which factors as $3(x-3)(x^2+3x+9)$ .

Key idea

Taking out the greatest common factor can reveal a difference of cubes inside an expression whose highest power is $x^4$ .
Problem 7: Mark complete
Problem 8 A forced middle coefficient

For some positive number $d$ and real constants $K,m$ , the identity $x^3+K=(x+d)(x^2+mx+9)$ holds for every real $x$ . Find $m$ , and name the two terms of the expanded right side that the identity forces to vanish once like terms are collected.
1. Hint 1
  Two expressions that agree for every real $x$ must have the same coefficient at each power of $x$ .
2. Hint 2
  Expand the right side, collect like terms, and compare coefficients with the left side power by power.
Answer

$m=-3$ ; the two terms that must vanish are the $x^2$ term and the $x$ term of the expansion.

Full solution

Expand the right side: its $x^2$ coefficient is $m+d$ , its $x$ coefficient is $9+dm$ , and its constant is $9d$ .
Since the two sides agree for every $x$ , and the left side has no $x^2$ term and no $x$ term, each of those two coefficients must be zero.
From $m+d=0$ comes $m=-d$ , and substituting that into $9+dm=0$ gives
$9-d^2=0.$
Since $d>0$ , this forces $d=3$ , so $m=-3$ .
The identity is then $x^3+27=(x+3)(x^2-3x+9)$ , a sum of cubes whose middle terms cancel on expansion.

Answer

$m=-3$ ; the two terms that must vanish are the $x^2$ term and the $x$ term of the expansion.

Key idea

Missing coefficients in a cube factorization are forced by the cancellation of both middle powers.
Problem 8: Mark complete
Problem 9 A proposed replacement

Let $a,b$ be positive real numbers. A student replaces the trinomial in $(a+b)(a^2-ab+b^2)$ with $(a-b)^2$ . Noor says this lowers the product by exactly $ab$ . Is Noor correct? Explain, and give the exact decrease.
1. Hint 1
  Compare the original trinomial with the expanded square before multiplying by the common binomial.
2. Hint 2
  The difference between the trinomials must also be multiplied by $a+b$ .
Answer

No; the decrease is $ab(a+b)$ , not $ab$ .

Full solution

The replacement square is $a^2-2ab+b^2$ .
The original trinomial exceeds it by $ab$ , since its middle term is $-ab$ rather than $-2ab$ .
Both trinomials are multiplied by the same factor $a+b$ , so the decrease in the product is that $ab$ multiplied by $a+b$ , which is
$ab(a+b).$
Noor compared the two trinomials but never multiplied by $a+b$ , so the claimed decrease is too small.
For $a=3$ and $b=2$ the products are $35$ and $5$ , a decrease of $30$ , while $ab$ is only $6$ .

Answer

No; the decrease is $ab(a+b)$ , not $ab$ .

Key idea

A change inside one factor is multiplied by the other factor, so compare the products, not just the trinomials.
Problem 9: Mark complete
Problem 10 Cubing a sum

Ana writes $(x+2)^3=x^3+8$ . Find every real $x$ for which her equation is true, and explain why it is not an identity.
1. Hint 1
  Cubing a sum is not the same as adding two cubes; the expansion of $(x+2)^3$ carries two middle terms.
2. Hint 2
  Bring both sides together and factor what is left, rather than dividing, so that no solution is lost.
Answer

$x=0$ and $x=-2$ ; the equation is not an identity.

Full solution

Expanding $(x+2)^3$ as $(x^2+4x+4)(x+2)$ and collecting terms gives
$x^3+6x^2+12x+8.$
Subtracting $x^3+8$ from both sides leaves
$6x^2+12x=0.$
Factoring the left side gives
$6x(x+2)=0,$
so $x=0$ or $x=-2$ .
Dividing by $x$ instead of factoring would have lost the solution $x=0$ .
At $x=0$ both sides are $8$ , and at $x=-2$ both sides are $0$ .
At $x=1$ the sides are $27$ and $9$ , so the equation fails there and cannot be an identity.

Answer

$x=0$ and $x=-2$ ; the equation is not an identity.

Key idea

A cube of a sum carries middle terms that a sum of cubes does not, so the two agree only at special inputs.
Problem 10: Mark complete

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