Question #10345

Prove that (x+y)3-(x-y)3-6y(x2-y2)=8y2
With solution.

Expert's answer

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

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

Substituting these formulae into the left part of initial equation we obtain

x^3+3x^2y+3xy^2+y^3-(x^3-3x^2y+3xy^2-y^3)-6y(x2-y2)=|...open the braces...| =

=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-6yx^2+6y^3=

=8y^2 so that's it

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

Substituting these formulae into the left part of initial equation we obtain

x^3+3x^2y+3xy^2+y^3-(x^3-3x^2y+3xy^2-y^3)-6y(x2-y2)=|...open the braces...| =

=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-6yx^2+6y^3=

=8y^2 so that's it

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