Question


Answer

Using the sum of cubes formula

(x-y)^3+(y-x-2)^3+8

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

=(x-y)[(x-y)^2-(y-x-2)^2+2(y-x-2)-4]

=(x-y)[(x-y-2)(x-y+2)-(y-x-2)^2+2(y-x-2)]

=(x-y)(y-x-2)(-x+y+2-y+x+2+2)

=6(x-y)(y-x-2)

Steven Zheng posted 1 year ago

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