Sum and Difference of Cubes: Additional Practice (Set 1)

Pull out the common factor $4$ first, then factor $x^3 - 8 = x^3 - 2^3$ as a difference of cubes.

$4x^3 - 32 = 4(x^3 - 8) = 4(x - 2)(x^2 + 2x + 4)$

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

The cube roots are $x$ and $2y$, since $8y^3 = (2y)^3$. This is a sum, so the middle sign is opposite the plus (minus).

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

The middle term is $ab = 2xy$ with a minus, and the last term $+4y^2$ is always positive.

A perfect cube has a whole-number cube root. Test the candidates.

$343 = 7^3$

Since $7 \times 7 \times 7 = 343$, it is a cube; $48$, $100$, and $200$ have no whole-number cube root.

The cube roots are $\sqrt[3]{27x^3} = 3x$ and $\sqrt[3]{64} = 4$, so $a = 3x$ and $b = 4$. This is a sum, so the middle sign is opposite the plus (minus).

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

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

Read the terms as $1^3 - (4x)^3$, a difference with $a = 1$ and $b = 4x$. The middle sign is opposite the minus (plus).

$1 - 64x^3 = (1 - 4x)(1 + 4x + 16x^2)$

The middle term $+4x$ is plus for a difference, and the last term $+16x^2$ is always positive.

Both terms must be perfect cubes. In $x^3 + 12$, the number $12$ is not a cube, since $2^3 = 8$ and $3^3 = 27$.

$x^3 + 12 \text{ is not a sum of cubes}$

The others are cubes: $x^3 + 1$, $8x^3 - 27 = (2x)^3 - 3^3$, and $64 - x^3 = 4^3 - x^3$.

Pull out the common factor $5$ first, then factor $x^3 + 8 = x^3 + 2^3$ as a sum of cubes.

$5x^3 + 40 = 5(x^3 + 8) = 5(x + 2)(x^2 - 2x + 4)$

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

Since $216 = 6^3$, the cube roots are $x$ and $6$. 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.

The 'O' in SOAP means the trinomial's middle term takes the opposite sign from the binomial. Here the binomial $a - b$ is minus, so the middle term is plus.

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

The binomial's sign is the 'Same', and the last term is 'Always Positive'.

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

$27x^3 + 1 = (3x + 1)(9x^2 - 3x + 1) = 0$

Setting $3x + 1 = 0$ gives $x = -\tfrac{1}{3}$. The trinomial $9x^2 - 3x + 1$ is always positive and never zero, so $x = -\tfrac{1}{3}$ is the only real solution.

The cube roots are $\sqrt[3]{1000x^3} = 10x$ and $1$, so $a = 10x$ and $b = 1$. This is a sum, so the middle sign is opposite the plus (minus).

$1000x^3 + 1 = (10x + 1)(100x^2 - 10x + 1)$

The first trinomial term is $(10x)^2 = 100x^2$, and the middle sign is minus for a sum.

For a sum of cubes the middle sign is opposite the binomial's plus, so it is minus, and the middle term is a single $ab$.

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

The forms with $2ab$ are perfect-square trinomials, and $a^2 + ab + b^2$ belongs to the difference of cubes.