Disturbing polynomials

Take third power of both side of x-2y=6...............(1)

That is

(x-2y)^3=6^3

x^3-8 y^3-3 x*2y(x-2y)=216

x^3-8y^3-6xy*6=216 (putting (1) value)

So

x^3-8y^3-36xy-216=0 (taking everything on left side)

So answer is 0

Its easy the solution......x^3-8y^3=x^3-(2y)^3 =(x-2y) (x^2+4y^2+2xy) =6(x^2+4y^2+2xy)=6x^2+24y^2+12xy as(x-2y=6) given so acc. to question:x^3 -8y^3- 36xy - 216=6x^2+24y^2+12xy- 36xy - 216= 6[x^2+4y^2-4xy-36]=6[(x+2y)^2-36]=6[36-36]=0

First find an expression with $x=2y+6$ as a root, you do it by rearranging the expression into:

$x-6=2y$

then we cube both sides:

${ (x-6) }^{ 3 }={ (2y) }^{ 3 }$

${ x }^{ 3 }-18{ x }^{ 2 }+108x-216=8{ y }^{ 3 }$

Make one side zero:

${ 8{ y }^{ 3 }-x }^{ 3 }+18{ x }^{ 2 }-108x+216=0$

since the above expression is zero, we can add it to the expression we're solving:

$\quad \quad (x^{ 3 }-{ 8y }^{ 3 }-36xy-216)\\ +\quad ({ 8{ y }^{ 3 }-x }^{ 3 }+18{ x }^{ 2 }-108x+216)\\ =\quad 18{ x }^{ 2 }-108x-36xy$

factor out $18x$

$18x(x-2y-6)$

notice that $x-6=2y\quad \Leftrightarrow \quad x-2y-6=0$

substitute that and you get the answer $0$

x^3-8y^3=x^3-(2y)^3 =(x-2y) (x^2+4y^2+2xy)
=6(x^2+4y^2+2xy)=6x^2+24y^2+12xy as(x-2y=6) given so acc. to question:x^3 -8y^3- 36xy - 216=6x^2+24y^2+12xy- 36xy - 216= 6[x^2+4y^2-4xy-36]=6[(x+2y)^2-36]=6[36-36]=0