Sum and Difference of Cubes: Practice

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

$2x^3 + 16 = 2(x^3 + 8) = 2(x + 2)(x^2 - 2x + 4)$

The middle sign is minus for a sum, and a single $ab = 2x$ (not $4x$) sits in the middle.

With $a = x$ and $b = y$, this is a difference of cubes. The binomial keeps the minus sign and the middle term takes the opposite (plus) sign.

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

The last term $+y^2$ is always positive, and the middle sign is plus for a difference.

The cube roots are $2x$ and $3y$, since $8x^3 = (2x)^3$ and $27y^3 = (3y)^3$. This is a sum, so the middle sign is opposite (minus).

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

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

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

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

The middle sign is minus for a sum, and the middle term is a single $ab = 2x$.

A difference of cubes needs both terms to be perfect cubes joined by a minus sign. Only $x^3 - 27$ fits, with $x^3 = (x)^3$ and $27 = 3^3$.

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

$5$ is not a cube, $x^2$ is a square, and $x^3 + 4x$ only shares a common factor of $x$.

Pull out the common factor $x^2$ first, then factor $x^3 - 27 = x^3 - 3^3$ as a difference of cubes.

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

The middle sign is plus for a difference, and the middle term is a single $ab = 3x$ (not $6x$).

The correct trinomial uses $ab = (x)(4) = 4x$, a single $ab$. The student used $8x = 2ab$, which is the perfect-square-trinomial pattern.

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

Here $x^2 + 8x + 16 = (x + 4)^2$, a different expression entirely.

The cube roots are $\sqrt[3]{64x^3} = 4x$ and $\sqrt[3]{125} = 5$, so $a = 4x$ and $b = 5$. This is a difference, so the middle sign is opposite the minus (plus).

$64x^3 - 125 = (4x - 5)(16x^2 + 20x + 25)$

The first trinomial term is $(4x)^2 = 16x^2$, and the middle sign is plus for a difference.

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

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

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

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

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

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

Since $1000 = 10^3$, the cube roots are $x$ and $10$. This is a difference, so the middle sign is opposite the minus (plus).

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

The middle term is a single $ab = 10x$ (not $20x$), and the last term $+100$ is always positive.

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

$8x^3 - 1 = (2x - 1)(4x^2 + 2x + 1) = 0$

Setting $2x - 1 = 0$ gives $x = \tfrac{1}{2}$. The trinomial $4x^2 + 2x + 1$ is always positive and never zero, so $x = \tfrac{1}{2}$ is the only real solution.