Question

Find the value of x3-8y3-3xy - 216whenx= 2y+6​

Answer 1

Solution:-x=2y+6

x^3-8y^3-3xy-216

therefore,(2y+6)^3-8y^3-3(2y+6)y-216

=(2y)^3+(6)^3+3(2y)(6)(2y-6)-8y^3-6y²-18y-216

=8y^3+216+36y(2y-6)-8y^3-6y²-18y-216

=8y^3+216+72y²-216y-8y^3-6y²-18y-216

=8y^3-8y^3+72y²-6y²-216y-18y+216-216

=66y²-234y ans

I hope it helps you

Plz.. mark brainliest