Question

Factor completely.

40x^3-5y^3

Cn you please show work so I can understand how to do it :)

Answer 1

40x^3-5y^3
= 5( 8x^3 - y^3 )
= 5 ( (2x)^3 - y^3 )

Now we use this identity:
http://www.mathsisfun.com/algebra/polynomials-difference-two-cubes.html

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

In our case:
a=2x
b=y

5( (2x)^3 - y^3 )
= 5 (2x-y)(4x^2+2xy+y^2)


More about factoring, and special identities:

http://en.wikipedia.org/wiki/Difference_between_two_cubes#Sum.2Fdifference_of_two_cubes