Proving that an equation has a solution 
If $a$ and $b$ are positive numbers, prove that the equation
  $$\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2}=0$$ 
  has at least one solution in the interval $(-1,1)$

My solution is as follows, I could not be sure on its correctness. If you have any other solutions, could you please share them?
My attempt: 
Let$$f(x)=\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2} = \frac{a}{(x+1)(x-(\frac{-1+\sqrt5}{2}))(x-(\frac{-1-\sqrt5}{2}))}+\frac{b}{(x-1)(x^2+x+2)}$$
Then, the two summands have distinct signs in the interval $(\frac{-1+\sqrt5}{2},1)$. Therefore, $f$ might take on the value zero in this interval.
We know that $f(x)\rightarrow-\infty$ as $x \rightarrow 1^-$ and $f(x)\rightarrow\infty$ as $x \rightarrow (\frac{-1+\sqrt5}{2})^+$.
Then, for a sufficiently small $\delta>0$, $f(1-\delta)<0$ and $f(\frac{-1+\sqrt5}{2}+\delta)>0$. Also, $f$ is continuous on $[\frac{-1+\sqrt5}{2}+\delta,1-\delta]$. Thus, from the Intermediate Value Theorem, there exist at least one $c\in(\frac{-1+\sqrt5}{2}+\delta,1-\delta)$ such that $f(c)=0$.

Sketch of solution. May be unstable. :-)
Rewrite $f(x)$ as
$$
f(x) = \frac{a(x^3+x-2)+b(x^3+2x^2-1)}{(x^3+2x^2-1)(x^3+x-2)}
$$
and denote the numerator by
$$
g(x) = (a+b)x^3+2bx^2+ax-(2a+b)
$$
and the denominator by
$$
h(x) = (x^3+2x^2-1)(x^3+x-2)
$$
Note that $h(x)$ has a zero (and therefore $f(x)$ has a pole) at $x_0 = \frac{\sqrt{5}-1}{2}$.  In this vicinity, $h(x)$ goes from positive to negative.
Now,
$$
g(x_0) = \frac{3\sqrt{5}-9}{2}
$$
which is a negative number, while $g(1) = 2b$, a positive number.  Therefore, in the interval $(x_0, 1)$, $f(x)$ goes from a large positive number to a large negative number.  Since there are no other poles in this interval, there must be a zero in that interval.
