<s>[INST] Find limits for uniformly distributed random variable given equations of expectations My question is this:
Let $\eta$ be uniformly distributed random variable over the interval $[a, b]$.
Let    E$\eta^2$ = 1 and E$\eta$ = -E$\eta^3$. Find a and b.
This is my approach:
$
  \eta\sim Unif[a,b]\Rightarrow
  f_\eta(x)=
   \dfrac{1}{b-a}, \qquad a\leq x\leq b\\~\\
 %
  E\eta\phantom{^2}=\int_{a}^{b} x\phantom{^2}f_\eta(x)dx= \int_{a}^{b} \dfrac{x}{b-a}dx=\left[ \dfrac{x^2}{2b-2a} \right]_{a}^{b}=\dfrac{b^2-a^2}{2b-2a}=\dfrac{b+a}{2}\\~\\~\\
  E\eta^2=\int_{a}^{b} x^2f_\eta(x)dx= \int_{a}^{b} \dfrac{x^2}{b-a}dx=\left[ \dfrac{x^3}{3b-3a} \right]_{a}^{b}=\dfrac{b^3-a^3}{3b-3a}=\dfrac{b^2+ab+a^2}{3}\\~\\~\\
  E\eta^3=\int_{a}^{b} x^3f_\eta(x)dx= \int_{a}^{b} \dfrac{x^3}{b-a}dx=\left[ \dfrac{x^4}{4b-4a} \right]_{a}^{b}=\dfrac{b^4-a^4}{4b-4a}=\dfrac{(b^2+a^2)(b+a)}{4}\\~\\~\\
  \begin{cases}
   \dfrac{b^2+ab+a^2}{3}=1\\~\\
   \dfrac{b+a}{2}=-\dfrac{(b^2+a^2)(b+a)}{4}
  \end{cases}\Rightarrow
   \begin{cases}
    b^2+ab+a^2=3\\~\\
    4(b+a)=-2(b^2+a^2)(b+a)
   \end{cases}\Rightarrow
    \begin{cases}
     b^2+ab+a^2=3\\~\\
     2b+2a=-b^3-b^2a-a^2b-a^3
    \end{cases}\Rightarrow
  $
This is where I am stuck. I can't solve this system of equations, but graphically I can see that $\pm\sqrt{3}$ is my solution. Can someone help me get to that solution or tell me what is wrong with my solution or maybe there is another approach?
 [/INST] We have $$4(b+a)=-2(b^2+a^2)(a+b)$$  If $b+a\neq0$ then $$b^2+a^2=-2$$ which is absurd.  Therefore, $$a=-b$$  Substitute this into $$\frac{b^2+ab+a^2}3=1$$
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