A problem related to $2x^3-3x^2-x+\frac{3}{2}=0$ 
let $f(x)=2x^3-3x^2-x+\frac{3}{2}$.Then prove that
$$\int_{1/8}^{7/8} f(f(x)) \text d x\neq \frac{3}{4}$$


Factorising$$f(x)=2(x-1.5)(x-\frac{1}{\sqrt{2}})(x+\frac{1}{\sqrt{2}})$$
$$f(f(x))=2(f(x)-1.5)(f^2(x)-0.5)$$
But i don't see anything nice from this.
Maybe if i could prove that $$f(f(x+\frac{1}{2}))=-f(f(x-\frac{1}{2}))$$ then the integral would turn out to be zero.
also i dont think we have to find the exact value of integral for second part if we could just set up an inequality or prove that the integral is negative we are done!
Please note that i am intersted in a proof without actually finding $f(f(x))$
 A: You have the right idea about exploiting some sort of symmetry in $f(f(x))$, but the integral doesn't evaluate zero.
Anyway, for all $y \in \mathbb{R}$, we have $$f(1-y) = 2(1-y)^3-3(1-y)^2-(1-y)+\dfrac{3}{2} = -2y^3+3y^2+y-\dfrac{1}{2} = 1-f(y).$$
Hence, for all $x \in \mathbb{R}$, we have$$f(f(1-x)) = f(1-f(x)) = 1-f(f(x)),$$ where we have used the above property for $y = x$, and then for $y = f(x)$.
Now, let $I = \displaystyle\int_{1/8}^{7/8}f(f(x))\,dx$. Replace $x$ with $1-x$ to get $I= \displaystyle\int_{1/8}^{7/8}f(f(1-x))\,dx$.
By adding these expressions for $I$ together, we get $$2I = \int_{1/8}^{7/8}f(f(x))+f(f(1-x))\,dx = \int_{1/8}^{7/8}1\,dx = \dfrac{3}{4},$$ and thus, $I = \dfrac{3}{8} \neq \dfrac{3}{4}$.
A: Too long for the comment.
Taking in account the brilliant condition
$$f(f(1-x)) = 1 - f(f(x))$$
by JimmyK4542, one can write
$$I=\int\limits_{\large^1/_8}^{\large^1/_2}f(f(x))\text dx +\int\limits_{\large^1/_2}^{\large^7/_8}f(f(x))\text dx 
=\int\limits_{\large^1/_8}^{\large^1/_2} \big(f(f(x))+f(f(1-x))\big)\text dx =\int\limits_{\large^1/_8}^{\large^1/_2}\text dx,$$
$$\color{brown}{\mathbf{I = \dfrac38.}}$$
