Question

Which long division problem can be used to prove the formula for factoring the difference between two perfect cubes?

• A. $a-b$ square root $a^2+ab+b^2$
• B. $a+b$ square root $a^2-ab+b^2$
• C. $a+b$ square root $a^3+0·a^2·b+0·a·b^2-b^3$
• D. $a-b$ square root $a^3+0·a^2·b+0·a·b^2-b^3$

Answer 1

The correct option is D

$a-b$ square root $a^3+0·a^2·b+0·a·b^2-b^3$

The explanation for the correct option:

Option (D):

The difference between the two perfect cubes may be represented by $a^3-b^3$.

It is known that $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\dots \left(1\right)$

Then, to prove divide $a^3-b^3$ by the first factor of the right side $a-b$.

Rewrite the $a^3-b^3$ as follows:

$a^3-b^3=a^3+0·a^2·b+0·a·b^2-b^3$

Thus, the equation $\left(1\right)$ can be written as follows:

$a^3+0·a^2·b+0·a·b^2-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$

By the division property of equality, divide both sides by the same factor, which in this case will be the binomial.

Thus, the equation is as follows:

$\frac{a^3+0·a^2·b+0·a·b^2-b^3}{\left(a-b\right)}=\left(a^2+ab+b^2\right)$

Hence, option (D) is correct.

The explanation for the incorrect options:

Options (A), (B), and (C):

The difference between the two perfect cubes may be represented by $a^3-b^3$.

So, the long division problem that can be used to prove the formula for factoring the difference between two cubes is:

$\frac{a^3+0·a^2·b+0·a·b^2-b^3}{\left(a-b\right)}=\left(a^2+ab+b^2\right)$

Therefore, options (A), (B), and (C) are incorrect.

Hence, the correct option is (D).