Question

if x=2y+6,find the value ofX^3-8Y^3-36XY-216

Answer 1

x=2y+6 .then x-2y-6=0
now value of x^3-8y^3-36xy-216=0
(x)^3-(2y)^3-(6)^6+3(x×2y×-6)=0
(x-2y-6)(x^2+4y^2+36-4xy+24y-12x)=0

putting the value of (x-2y-6)=0
then
0×(x^2+4y^2+36-4xy+24y-12x)=0 ans.

Answer 2

x=2y +6 then x^3-8y ^3-36xy-216
=(2y+6)^3 -8y^3-36y(2y+6)-216
=(2y)^3+(6)^3+3.(2y)^2.6+3.2y. (6)^2 -8y^3 -72y^2-216y-216
=8y^3 +216 +3.4y^2.6+6y.36 -8y^3 -72y^2-216y-216
=72y^2+216y-72y^2-216y
=0
hope it will help u