How do u derive the difference of two perfect cubes?

Answer 1

Is this a factoring question?
(x^3 - y^3)=(x - y)(x^2 + 2xy + y^2)

Answer 2

Difference Of Two Perfect Cubes

Answer 3

(EDIT; mathchick, your second factor is wrong)

How do you DERIVE it or VERIFY it?

If you want to VERIFY that x^3 - y^3 factors into (x-y)(x^2 + xy + y^2) , then you just distribute a bunch. Multiply x times (x^2 + xy + y^2), then multiply -y times (x^2 + xy + y^2), and gather up your results.

If you want to DERIVE it, well, that's different.

Hmmm, lemme think....



Ok. Let x^3 - y^3 = 0. Since x = y makes this equaiton true, then (x - y) must be a factor of x^3 - y^3. With that knowledge, you can do long division with x^3 - y^3 divided by x - y (or even synthetic division if you are careful). Dividing will give you a quotient of x^2 + xy + y^2. Rewrite the division problem as a multiplication problem and you get

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

Note: the statement about x = y being a solution to x^3 - y^3 = 0 and leading to the factor of (x - y) is actually unnecessary. Just dividing x^3 - y^3 by x - y and getting a quotient of x^2 + xy + y^2 is enough to "derive" the formula. The trick is kowing that (x - y) was a factor before the division proiblem gave a remainder of zero, and thus knowing to use that particular dividend.

Answer 4

Difference of Cubes - An expression of the form a3 - b3. The difference of two cubes factors into (a - b)(a2 + ab + b2).

Answer 5

let x be the side of one cube and n be the diff in length

difference = x^3 - (x-n)^3

= x^3 - (x^3 - 3nx^2 + 3n^2x - 3n^3)
= 3nx^2 - 3n^2 + 3n^3
= 3n(x^2-n+n^2)