Sum and Difference of Cubes: Practice

Take the cube root of each term: $x^3 = (x)^3$ and $8 = 2^3$, so $a = x$ and $b = 2$. This is a sum of cubes, so the binomial keeps the plus sign and the trinomial's middle term takes the opposite (minus) sign.

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

The middle term must be minus for a sum, so $x^2 + 2x + 4$ is wrong.

Both terms are perfect cubes, since $x^3 = (x)^3$ and $64 = 4^3$, and they are added.

$x^3 + 64 = x^3 + 4^3$

Two cubes joined by a plus sign form a sum of cubes.

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

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

The middle term is $-x$, and $(x + 1)^3$ is a different expression with extra terms.

The cube roots are $x$ and $1$, so $a = x$ and $b = 1$. This is a difference, so the binomial keeps the minus sign and the middle term takes the opposite (plus) sign.

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

The middle term must be plus for a difference, so $x^2 - x + 1$ is wrong.

A difference of cubes is two perfect-cube terms joined by a minus sign. Only $x^3 - 8$ fits, with $x^3 = (x)^3$ and $8 = 2^3$.

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

$x^3 + 8$ is a sum, $x^2$ is a square not a cube, and $7$ is not a perfect cube.

Look for the whole number whose cube is $64$.

$4^3 = 4 \times 4 \times 4 = 64$

So $\sqrt[3]{64} = 4$. (Note that $8^2 = 64$, but $8$ is the square root, not the cube root.)

For a difference of cubes, the binomial is $a - b$ and the trinomial has an opposite (plus) middle sign with a positive last term.

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

The middle term is $+ab$ and the last term is $+b^2$.

To factor $x^2 + 4x + 16$ you would need two integers that multiply to $16$ and add to $4$. The pairs multiplying to $16$ add to $17$, $10$, and $8$, never $4$.

$x^2 + 4x + 16 \text{ does not factor over the integers}$

The cube trinomial always stops here; $(x + 4)^2 = x^2 + 8x + 16$, which is a different expression.

The letters SOAP stand for Same, Opposite, Always Positive. The last part describes the trinomial's final term.

$a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$

The last term $b^2$ is a square, and it is always written positive.

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

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

The middle sign is minus for a sum, so $x^2 + 3x + 9$ is wrong.

A perfect cube has a whole-number cube root. Check each: $8 = 2^3$, $125 = 5^3$, and $1 = 1^3$.

$50 \text{ has no whole-number cube root}$

Since $3^3 = 27$ and $4^3 = 64$, there is no whole number that cubes to $50$.

The 'S' in SOAP means the binomial takes the same sign as the original expression. A sum of cubes gives $(a + b)$; a difference gives $(a - b)$.

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

The plus in $a^3 + b^3$ carries straight into the binomial $a + b$.