Question

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

Answer 1

Answer:

The correct options, rearranged, are:

Options:

$A)(a^2+ab+b^2)/(a-b)\\\\B)(a^2-ab+b^2)/(a+b)\\\\C)(a^3+0a^2+0ab^2-b^3)/(a+b))\\\\ D)(a^3+0a^2+0ab^2-b^3)/(a-b)$

And the asnwer is the last option (D).

Explanation:

You need to find which long division can be used to prove the formula for factoring the difference of two perfect cubes.

The difference of two perfect cubes may be represented by:

• $a^3-b^3$

And it is, as a very well known special case:

• $a^3-b^3=(a-b)(a^2+ab+b^2)$

Then, to prove, it you must divide the left side,    $a^3-b^3$     , by the first factor of the right side,    $a-b$

Note that, to preserve the places of each term, you can write:

• $(a^3-b^3)=(a^3+0a^2+0ab^2-b^3)$

Then, you have:

• $(a^3+0a^2+0ab^2-b^3)=(a-b)(a^2+ab+b^2)$

By the division property of equality, you can divide both sides by the same factor, which in this case will be the binomial, and you get:

• $(a^3+0a^2+0ab^2-b^3)/(a-b)=(a^2+ab+b^2)$

That is the last option (D).

Answer 2

Some of the possible options of the questions are;

A) $(a - b) | \overline {a^2 + a \cdot b + b^2}$

B) $(a + b) | \overline {a^2 - a \cdot b + b^2}$

C) $(a + b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

The difference of two perfect cubes has a binomial factor and a trinomial factor

The option that gives the long division problem that can be used to prove the difference of two perfect cubes is option D

D) $\underline {(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}}$

Reason:

The formula for factoring the difference of two perfect cubes is presented as follows;

a³ - b³ = (a - b)·(a² + a·b + b²)

Given that a factor of the difference of two cubes is (a - b), and that we

have; (a³ + 0·a·b² + 0·a²·b - b³) = (a³ - b³), both of which are present in

option D, by long division of option D, we have;

${} \hspace {33} a^2 + a \cdot b + b^2\\(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a^2 \cdot b - b^3}\\{} \hspace {33} \underline{a^3 - a^2 \cdot b }\\{} \hspace {55} a^2 \cdot b + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\\ {} \hspace {55} \underline{a^2 \cdot b - a \cdot b^2}\\{} \hspace {89} a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3\\{} \hspace {89} \underline{a \cdot b^2 - b^3}\\{}\hspace {89} 0$

By the above long division, we have;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = a² + a·b + b²

Which gives;

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$ = (a³ + 0·a·b² + 0·a·b² - b³)/(a - b)

We get;

(a³ + 0·a·b² + 0·a·b² - b³)/(a - b) = a² + a·b + b²

(a - b)·(a² + a·b + b²) = (a³ + 0·a·b² + 0·a·b² - b³) = (a³ - b³)

(a - b)·(a² + a·b + b²) = (a³ - b³)

(a³ - b³) = (a - b)·(a² + a·b + b²)

Therefore;

The long division problem that can be used to prove the formula for

factoring the difference of two perfect cubes is

$(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$, which is option D

D) $(a - b) | \overline {a^3 + 0 \cdot a \cdot b^2 + 0 \cdot a \cdot b^2 - b^3}$

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