<s>[INST] 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?
 [/INST] To solve this problem, first of all think about the x values for which the following function is undefined.
We have,
$$f(x) = \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}$$
which can be written as,
$$f(x) = \frac{a}{(x+1)(x^2+x-1)} + \frac{b}{(x-1)(x^2+x+2)}$$ 
Note that the above function is undefined for
$$x = \pm 1\ , \ \frac{-1\pm \sqrt{5}}{2} \ ...(1)$$
Our interval is $(-1,1)$, so forget about $\frac{-1 - \sqrt{5}}{2}$ which is approximately $-1.62$.
Now, let's calculate the following limits,
$$\lim_{x \to 1^{-}}f(x) =\lim_{x \to 1^{-}}\left( \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}\right) = -\infty $$ as the first term is simply $\frac{a}{2}$ and the function $x^3 + x - 2$ goes to $0$ from behind as x approaches $1$ from the left. So, the second term in the function is a huge negative number.
And,
$$\lim_{x \to {\left( \frac{-1 + \sqrt{5}}{2}\right)}^{+}}f(x) =\lim_{x \to {\left( \frac{-1 + \sqrt{5}}{2}\right)}^{+}}\left( \frac{a}{x^3 + 2x^2 - 1} + \frac{b}{x^3 + x - 2}\right) = \infty $$ as the function $x^3 + 2x^2 - 1$ approaches $0$ from above from x approaches $\frac{-1 + \sqrt{5}}{2}$ from the right. So, the first term is actually a big positive number and the second term which is actually defined is negligible.
Now, the reason for calculating these limits was to get to the answer by saying that between x = $\frac{-1 + \sqrt{5}}{2}$ and x = $1$, $f(x)$ is continuous everywhere (see $(1)$). The limits for these values of x show that the function has to cross the x-axis between these values of x, that is, there should be at least one root!
The given graph for $ \frac{2}{x^3 + 2x^2 - 1} + \frac{3}{x^3 + x - 2}$
  illustrates the above proof:

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