Sum and Difference of Cubes: Practice

The cube roots are $\sqrt[3]{27x^3} = 3x$ and $1$, so $a = 3x$ and $b = 1$. This is a sum, so the middle sign is opposite the plus.

$27x^3 + 1 = (3x + 1)(9x^2 - 3x + 1)$

The first trinomial term is $(3x)^2 = 9x^2$, and the middle term is minus for a sum.

Take the cube root of each term. Since $64x^3 = (4x)^3$ and $27 = 3^3$, the roots give $a$ and $b$.

$64x^3 - 27 = (4x)^3 - 3^3, \quad a = 4x, \ b = 3$

The coefficient root is $\sqrt[3]{64} = 4$, not $8$ (that is $\sqrt{64}$).

A sum of cubes factors with an opposite (minus) middle term. Expanding confirms the middle terms cancel.

$(x + 2)(x^2 - 2x + 4) = x^3 + 8$

The form $(x + 2)^3 = x^3 + 6x^2 + 12x + 8$ has extra terms, and a plus middle term does not cancel.

Since $216 = 6^3$, the cube roots are $x$ and $6$, so $a = x$ and $b = 6$. This is a sum, so the middle sign is opposite (minus).

$x^3 + 216 = (x + 6)(x^2 - 6x + 36)$

The last term $+36 = 6^2$ is always positive, and the middle sign is minus for a sum.

Both terms must be perfect cubes. In $8x^3 + 27$, $8x^3 = (2x)^3$ and $27 = 3^3$.

$8x^3 + 27 = (2x)^3 + 3^3$

$8x^2$ is not a cube, $20$ is not a cube, and $9$ is not a perfect cube, so the others fail.

Since $1000 = 10^3$, read the terms as $10^3 - x^3$, a difference with $a = 10$ and $b = x$. The middle sign is opposite the minus (plus).

$1000 - x^3 = (10 - x)(100 + 10x + x^2)$

The first trinomial term is $10^2 = 100$, so a $10$ there is wrong, and the middle sign is plus.

For a sum of cubes the middle sign is opposite the binomial's plus, so it must be minus.

$x^3 + 8 = (x + 2)(x^2 - 2x + 4)$

The binomial $x + 2$ and the last term $+4$ are correct; only the middle sign was flipped.

Read the terms as $4^3 - x^3$, a difference with $a = 4$ and $b = x$. The trinomial is $a^2 + ab + b^2$.

$64 - x^3 = (4 - x)(16 + 4x + x^2)$

The first term is $4^2 = 16$, the middle is $ab = 4x$ (not $8x$), and the last is $x^2$.

For $x^3 + 27$ the correct trinomial is $x^2 - 3x + 9$, using $ab = 3x$. Doubling it to $2ab = 6x$ gives the perfect-square-trinomial mistake.

$x^2 - 6x + 9 = (x - 3)^2 \ne x^2 - 3x + 9$

The cube trinomial never doubles the middle term, so $x^2 - 6x + 9$ is the wrong one.

The cube roots are $\sqrt[3]{125x^3} = 5x$ and $\sqrt[3]{8} = 2$, so $a = 5x$ and $b = 2$. This is a sum, so the middle sign is opposite the plus (minus).

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

The first trinomial term is $(5x)^2 = 25x^2$, and the middle sign is minus for a sum.

A sum of cubes is not the cube of a sum. Expanding $(x + 3)^3$ produces middle terms that $x^3 + 27$ does not have.

$(x + 3)^3 = x^3 + 9x^2 + 27x + 27$

The correct factorization is $x^3 + 27 = (x + 3)(x^2 - 3x + 9)$, a binomial times a trinomial.

Since $216 = 6^3$, the cube roots are $x$ and $6$, so $a = x$ and $b = 6$. This is a difference, so the middle sign is opposite the minus (plus).

$x^3 - 216 = (x - 6)(x^2 + 6x + 36)$

The middle sign is plus for a difference, and the last term $+36$ is always positive.