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?

Proof:
Let $f(x)=\frac{a}{x^{3}+2 x^{2}-1}+\frac{b}{x^{3}+x-2} .$ We want to prove that $f$ has at least one solution in the interval
$(-1,1) .$ It suffices to prove that $f(x)=0 .$ By factorization, $f(x)=\frac{a}{(x+1)(x-a)(x-b)}+$$\frac{b}{(x-1)\left(x^{2}+x+2\right)} .$ We denote $\frac{-1+\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}$ as $a, b$ respectively. We find $x$ such that $f$ is discontinuous. Clearly, $x=\pm 1, a, b .$ However, since the given interval is $(-1,1)$ then $f$ is discontinuous at $x=a .$ Now, suppose $r, s$ is in $(a, 1)$ with $r>a$ and $s<1$ Clearly, $f$ is continuous on $[r, s] .$ Note that, $\lim _{x \rightarrow a^{+}} f(x)=\infty$ and $\lim _{x \rightarrow 1^{-}} f(x)=-\infty .$ Hence, by the Intermediate Value Theorem, since $f$ is continuous on the closed interval $[r, s]$ and previously we have shown that $-\infty<f(x)<\infty$ then there exist a $c \in(r, s)$ such that $f(c)=0$. Hence,
$f(x)$ has at least one solution in $(-1,1)$ namely, $x=c$.
