Area under the graph - integration 
The region $P$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the line $x=a$ . The region $Q$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the lines $x=2a$ and $x=a$. Given that the area of $Q$ is twice the area of $P$, find the value of $a$ .


Firstly , on the first step , in already stuck ...
I used definite integral to find the area of $P$ -
$$\int^a_0\ (3x-x^2)dx=\frac{9a^2-2a^3}{6}$$
However when I calculate area of $Q$ , it's the same as Area of $P$ - $$\frac{9a^2-2a^3}{6}$$
Then since
$Q= 2P$
$9a^2 - 2a^3 = 18a^2 - 4a^3 $
From here, I definitely can't find the value of $a$ ... where have I gone wrong or misunderstood ?

Since $a\neq 0$ (as this would be absurd), we can solve your equation as follows 
\begin{align}9a^2−2a^3&=18a^2−4a^3\\
9-2a&=18-4a\tag{divide through by $a$}\\
2a&=9\\
a&=\frac 92\end{align}
However, as noted in other answers, this is incorrect. 
Your error is somewhere in your calculation of the area of $Q$, you should get $$\int_a^{2a}(3x-x^2)\text dx =\frac{27a^2-14a^3}{6}$$
This then gives you \begin{align}Q&=2P\\\frac{27a^2-14a^3}{6}&=2\times \frac{9a^2-2a^3}{6}\\
27a^2-14a^3&=2(9a^2-2a^3)\\
27a^2-14a^3&=18a^2-4a^3\\
27-14a&=18-4a\tag{divide through by $a$}\\
9&=10a\\
a&=\frac 9{10}\end{align}
