pls solve this easy question
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x=2y+6..........given
x-2y-6=0
we know that if x+y+z=0 then
x^3+y^3+z^3-3xyz=0
(x)^3-8y^3-216-3×x×(-2y)(-6)=0
x^3-8y^3-216-36xy=0
hence the value is 0
x-2y-6=0
we know that if x+y+z=0 then
x^3+y^3+z^3-3xyz=0
(x)^3-8y^3-216-3×x×(-2y)(-6)=0
x^3-8y^3-216-36xy=0
hence the value is 0
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