Trying to derive expected value of triangularly distributed random variable I'm having trouble deriving the expected value of a triangularly distributed random variable with lower bound $a$ upper bound $b$ and mode $c$ for the case when the distribution is symmetric about the mode. The expected value in this case is known to be the mode $c$. The density function of the triangular distribution can be found here. In the symmetric case, this reduces to
$$ P(x) = 
     \begin{cases}
       \frac{x-a}{\beta^2} & a \leq x < c \\[8pt]
      \frac{-(x-b)}{\beta^2} & c \leq x \leq b
     \end{cases}
$$
where the parameter $\beta$ is defined by
$$a = c-\beta$$
$$b = c + \beta$$
The expected value integral is then
$$E(x)=1/\beta^2\left(\int_a^cx(x-a)\:dx+\int_c^b -x(x-b)\:dx\right)$$
Which evaluates to
$$E(x)=1/\beta^2\left(\left[\frac{x^3}{3}-\frac{ax^2}{2}\right]^{x=c}_{x=a}-\left[\frac{x^3}{3}-\frac{bx^2}{2}\right]^{x=b}_{x=c}\right)$$
$$E(x)=1/\beta^2\left(\frac{c^3}{3}-\frac{ac^2}{2}+\frac{a^3}{6}-\left(-\frac{b^3}{6} - \frac{c^3}{3}+\frac{bc^2}{2}\right)\right)$$
$$E(x)=1/\beta^2\left(\frac{2c^3}{3}+\frac{a^3}{6}+\frac{b^3}{6}-\frac{ac^2}{2} -\frac{bc^2}{2}\right)$$
$$E(x)=\frac{1}{6\beta^2}\left(4c^3+a^3+b^3-3c^2(a+b)\right)$$
$$E(x)=\frac{1}{6\beta^2}\left(-2c^3+a^3+b^3\right)$$
Here is where I throw my hands in the air and turn to MSE. How the heck is this going to reduce to $c$??

UPDATE
Finally you got
$$E(x)=\frac{1}{6\beta^2}\left(-2c^3+a^3+b^3\right)$$
but notice that since $a=c-\beta$ and $b=c+\beta$, we have
$$
\begin{split}
a^3+b^3-2c^3
   &=(c-\beta)^3 + (c+\beta)^3-2c^3\\
   &= \left(c^3-3c^2\beta+3c\beta^2-\beta^3\right)
      + \left(c^3+3c^2\beta+3c\beta^2+\beta^3\right)
      - 2c^3\\
   &= 3c\beta^2+3c\beta^2 \\
   &= 6c\beta^2
\end{split}
$$
Thus
$$
E(x)=\frac{-2c^3+a^3+b^3}{6\beta^2} = \frac{6c\beta^2}{6\beta^2} = c,
$$
as desired.
