Factor $64x^9 - 125y^6$

Question

I'm trying to help my daughter with Algebra 2. This is a homework assignment. I've done a fair amount of searching but I just can't figure this out. Mathway, an online tool gave the answer:

$(4x^3 - 5y^2) (16x^6 + 20x^3y^2 + 25y^4)$

I can pattern match a bit but can't figure out what's going on. Would someone be willing to provide a link or explain what sort of procedure to follow to factor this expression?

Answer 1

You should see the original expression as the difference of two cubes. The factorization you have implements $a^3-b^3=(a-b)(a^2+ab+b^2)$

Answer 2

One starts by observing that $64x^9 - 125y^6 = (4x^3)^3 - (5y^2)^3$.

Answer 3

Way to do this:

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

$64x^9 - 125y^6 = (4x^3)^3 - (5y^2)^3$

$= (4x^3 - 5y^2) [(4x^3)^2 + 4x^3 * 5y^2 + (5y^2)^2]$

$= (4x^3 - 5y^2) (16x^6 + 20x^3y^2 + 25y^4)$