Answer
The equation $~~\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2} = 0~~$ has at least one solution in the interval $(-1,1)$
Work Step by Step
$\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2} = 0$
$\frac{a(x^3+x-2)+b(x^3+2x^2-1)}{(x^3+2x^2-1)(x^3+x-2)} = 0$
This expression is equal to 0 when the numerator is equal to 0.
We can evaluate the numerator when $x=1$:
$a[(1)^3+(1)-2]+b[(1)^3+2(1)^2-1] = a(0) +b(2) = 2b \gt 0$
We can evaluate the numerator when $x=-1$:
$a[(-1)^3+(-1)-2]+b[(-1)^3+2(-1)^2-1] = a(-4) +b(0) = -4a \lt 0$
Note that the expression for the numerator $~~a(x^3+x-2)+b(x^3+2x^2-1)~~$ is continuous on the interval $(-1,1)$
According to the Intermediate Value Theorem, there is a number $c$ in the interval $(-1,1)$ such that the numerator $~~a(c^3+c-2)+b(c^3+2c^2-1) = 0$
Therefore, the equation $\frac{a}{x^3+2x^2-1}+\frac{b}{x^3+x-2} = 0$ has at least one solution in the interval $(-1,1)$
