Sketch the curve with equation

$y=x^2(2x-a)$

where $a>0$.

The polynomial $p(x)$ is given by

$p(x)=x^2(2x-a)-64$

It is given that $x-8$ is a factor of $p(x)$. Use the factor theorem to show that $a=15$.

State the transformation which maps the curve with equation $y=x^2(2x-15)$ onto the curve with equation $y=x^2(2x-15)-64$.

The polynomial $x^2(2x-15)-64$ can be written as $(x-8)(2x^2+bx+c)$. Without finding the values of $b$ and $c$, use your knowledge of the transformation in part (b)(ii) and the sketch in part (a) to explain why

$b^2<8c$