Hi
Didn't you understand the previous clarification ?
http://www.mathhelpforum.com/math-he...ed-powers.html
$\displaystyle 2(x^2-6)^9+(x^2-6)^4+4(x^2-6)^3+(x^2-6)^2$
let $\displaystyle u = x^2-6$
$\displaystyle 2u^9+u^4+4u^3+u^2$
factor out $\displaystyle u^2$ ...
$\displaystyle u^2(2u^7 + u^2 + 4u + 1)$ ... see the "1" ?
back substitute ...
$\displaystyle (x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4(x^2-6) + 1]
$
distribute the $\displaystyle 4(x^2-6)$ term ...
$\displaystyle (x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4x^2 - 24 + 1]$
$\displaystyle (x^2-6)^2[2(x^2-6)^7 + (x^2-6)^2 + 4x^2 - 23]$
$\displaystyle (15xy-1)(2x-1)^3 - 8(2x-1)^2$
let $\displaystyle u = (2x-1) $
$\displaystyle (15xy-1)u^3 - 8u^2$
factor out $\displaystyle u^2$ ...
$\displaystyle u^2[(15xy-1)u - 8]$
back substitute $\displaystyle (2x-1)$ for $\displaystyle u$ ...
$\displaystyle (2x-1)^2[(15xy-1)(2x-1) - 8]$
note that it is permissible to stop here, or one can expand the binomial product in the brackets and combine the resulting constant terms ...
$\displaystyle (2x-1)^2[(30x^2y - 15xy - 2x + 1) - 8]
$
$\displaystyle (2x-1)^2(30x^2y - 15xy - 2x - 7)$
really no big advantage in doing this, however.