Sum and Difference of Cubes: Additional Practice (Set 2)

Pull out the common factor $3$ first, then factor the difference of cubes $x^3 - 8 = x^3 - 2^3$.

$3x^3 - 24 = 3(x^3 - 8) = 3(x - 2)(x^2 + 2x + 4)$

The binomial keeps the minus sign, and the middle term is plus for a difference.

The cube roots are $x$ and $\sqrt[3]{64} = 4$, so $a = x$ and $b = 4$. This is a sum, so the middle sign is opposite the plus (minus).

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

The middle sign is minus for a sum, and $(x + 4)^3$ has extra middle terms.

Factor as a difference of cubes, then use the zero-product property.

$x^3 - 8 = (x - 2)(x^2 + 2x + 4) = 0$

Setting $x - 2 = 0$ gives $x = 2$. The trinomial $x^2 + 2x + 4$ is always positive and never zero, so $x = 2$ is the only real solution.

The cube roots are $\sqrt[3]{125x^3} = 5x$ and $\sqrt[3]{27} = 3$, so $a = 5x$ and $b = 3$. This is a difference, so the middle sign is opposite the minus (plus).

$125x^3 - 27 = (5x - 3)(25x^2 + 15x + 9)$

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

Read the terms as $3^3 + (2x)^3$, a sum of cubes with $a = 3$ and $b = 2x$. The middle sign is opposite the plus (minus).

$27 + 8x^3 = (3 + 2x)(9 - 6x + 4x^2)$

The first trinomial term is $3^2 = 9$, the middle is $ab = 6x$ with a minus, and the last is $(2x)^2 = 4x^2$.

Factor as a difference of cubes, then use the zero-product property.

$x^3 - 125 = (x - 5)(x^2 + 5x + 25) = 0$

Setting $x - 5 = 0$ gives $x = 5$. The trinomial $x^2 + 5x + 25$ is always positive and never zero, so $x = 5$ is the only real solution.

Factor as a sum of cubes, then use the zero-product property.

$x^3 + 27 = (x + 3)(x^2 - 3x + 9) = 0$

Setting $x + 3 = 0$ gives $x = -3$. The trinomial $x^2 - 3x + 9$ is always positive and never zero, so $x = -3$ is the only real solution.

Pull out the common factor $2x$ first, then factor $x^3 + 64 = x^3 + 4^3$ as a sum of cubes.

$2x^4 + 128x = 2x(x^3 + 64) = 2x(x + 4)(x^2 - 4x + 16)$

The middle sign is minus for a sum, and the middle term is a single $ab = 4x$ (not $8x$).

Both a sum and a difference of cubes factor, and each gives a binomial times a trinomial.

$a^3 + b^3 = (a + b)(a^2 - ab + b^2), \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Unlike a sum of squares, a sum of cubes does factor, so 'only the difference factors' is false.

For a difference of cubes the middle sign is opposite the binomial's minus, so it is plus, and the middle term is a single $ab = (x)(3) = 3x$.

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

A $+6x$ would be $2ab$, the perfect-square mistake, not the cube pattern.

Pull out the common factor $2$ first, then factor $27x^3 + 8 = (3x)^3 + 2^3$ as a sum of cubes.

$54x^3 + 16 = 2(27x^3 + 8) = 2(3x + 2)(9x^2 - 6x + 4)$

The binomial keeps the plus sign, and the middle term is minus for a sum.

Read the terms as $2^3 - (3x)^3$, a difference with $a = 2$ and $b = 3x$. The middle sign is opposite the minus (plus).

$8 - 27x^3 = (2 - 3x)(4 + 6x + 9x^2)$

The first trinomial term is $2^2 = 4$, the middle is $ab = 6x$ with a plus, and the last is $(3x)^2 = 9x^2$.