$\displaystyle \displaystyle y=(1-x^2)^3(6+2x)^{-3}$

What I did was:

$\displaystyle \displaystyle=\frac{(1-x^2)^3}{(6+2x)^3}$

$\displaystyle \displaystyle=\frac{-6x(6+2x)^3(1-x^2)^2-6(1-x^2)^3(6+2x)^2}{(6+2x)^6}$

$\displaystyle \displaystyle=\frac{(1-x^2)^2(6+2x)^2(6x^2-12x-48)}{6+2x)^6}$

$\displaystyle \displaystyle=\frac{6(1-x^2)^2(x^2-2x-8)}{(6+2x)^4}$

While the answer in the back of the book is:

$\displaystyle \displaystyle=\frac{-6(1-x^2)^2(x^2+6x+1)}{(6+2x)^4}$

Do I have a calculation error or is the way I handled it wrong? Thanks!