Prove $ \ \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2} \ = \ 0 \ $ has a solution in $ \ (-1,1) $ 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) \ $ .
The question is from the exercises section of a textbook chapter on limits/continuity.
I've been stumped on this one for a couple of days. I've been trying to calculate $\lim _{x \to -1}$ and $\lim _{x \to 1}$ and then show the function is continuous to show a root must lie in the interval. Factorising the denominators gives...
$$\frac{a}{(x+1)(x^2+x-1)} + \frac{b}{(x-1)(x^2+x+2)} = 0$$
So of course $x = 1$ and $x = -1$ are undefined and so the limits will be one-sided. Playing around with equation I haven't been able to find an equivalent function across $x \neq -1, x \neq 1$.
The only thing I have been able to show is
$$\frac{a}{b} = - \frac{(x+1)(x^2+x-1)}{(x-1)(x^2+x+2)}$$
and so
$$\lim _{x \to -1} \frac{a}{b} = 0, \lim _{x \to 1} \frac{b}{a} = 0$$
but I'm not sure if this is significant or I'm overthinking things. Could anyone point me in the right direction?

In the interval $(-1,1)$ your equation is equivalent to 
$f(x):= a(x-1)(x^2+x+2)+b(x+1)(x^2+x-1)=0$
Now you can see that $f(x)$ is continuous and 
$\lim\limits_{x\to -1}f(x)<0, \quad 
\lim\limits_{x\to 1}f(x)>0$
Thus $f(x)=0$ has at least one solution (in the interval $(-1,1)$).
