Question #10345
Prove that (x+y)3-(x-y)3-6y(x2-y2)=8y2
With solution.
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
