Question

Can someone please show me step by step, how to factor this equation? a^3x-b^3x-a^3y+b^3y

the answer = (a-b)(a^2+ab+b^2)(x-y)

Answer 1

a^3x-b^3x-a^3y+b^3y
(a^3-b^3)x - (a^3 -b^3)y

Answer 2

Did u get that step?

Answer 3

okay, yes. First you take the x and y out

Answer 4

what next?

Answer 5

One could also start by taking a and b out:\[a^{3}(x-y)+b^{3}(y-x)\] But I don't really know how to get to that answer or why to aim at that? :o

Answer 6

The best way to do it:
FIND THE COMMON NUMBER BETWEEN EVERYTHING.

Answer 7

\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 8

Got the answer?

Answer 9

You can check that what abdul said by going backwards:
\[(a - b)(a ^{2} + ab + b ^{2})\]
\[= (a^{3}+a^{2}b + ab^{2}) + (-a^{2}b - ab^{2} - b^{3})\]
\[= a^{3}+a^{2}b - a^{2}b + ab^{2} - ab^{2} - b^{3}\]
\[= a^{3} - b^{3}\]
Not sure yet how to get that (x - y) out o.O

Answer 10

\[a^3x-b^3x-a^3y+b^3y\]
\[x(a^3-b^3)-y(a^3-b^3)\]
\[(a^3-b^3)(x-y)\]
\[(a-b)(a^2+ab+b^2)(x-y)\]