Can't find the polynomial root

Question

For which values ​​of the parameter $m$, square trinomial $f(x)=mx^2+(m+1)x+1$ has two different roots such that:

$x_1^3-x_2^3=x_1^4-x_2^4$ (the value of the parameter $m$ should be approximated).

After calculating this condition (using Viete) $x_1^3-x_2^3=x_1^4-x_2^4$, I've received this equation: $2m^3+2m^2+2m+1=0$. The problem is I don't know how to find the polynomial roots of this equation.I know that it should be $\approx-0.65$, but I don't know how to solve it. Thank you in advance.

Answer 1

$$2m^3+2m^2+2m+1=0$$

Following the steps given here, you have $\Delta=-44$; so, there is only one real root.

Using the hyperbolic method, you will obtain $$m=-\frac{1}{3} \left(1+2 \sqrt{2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{13}{8\sqrt{2}}\right)\right)\right)=-0.647799\cdots$$

Answer 2

HINTS:

$$1 + x_1 + x_1^2 = (1+x_2)(1+x_2^2)$$

$$ x_1+x_2= \frac{-(m+1)}{m}$$

$$ x_1 x_2= \frac{1}{m}$$