I have
(y^2/y+3)^4
The answer I keep getting is
(4y^8+6y)/(y+3)^5
But i keep getting told that's wrong so I'm not sure where I messed up.
$\displaystyle
\left(\frac{y^2}{y+3}\right)^4 = 4\left(\frac{y^2}{y+3}\right)^3\left(\frac{y^2}{y+ 3}\right)' =
$
$\displaystyle
4 \left( \frac{y^6}{(y+3)^3}\right) \left( \frac{2y(y+3)-y^2}{(y+3)^2}\right)
$
$\displaystyle
4 \left( \frac{y^6}{(y+3)^3}\right) \left( \frac{y^2+6y}{(y+3)^2}\right)
$
$\displaystyle
\frac{4y^7(y+6)}{(y+3)^5}
$