Solving Cubic root equation

Question

This is the equation I have $$ [\frac{k^3\alpha^2x_{N-1}^2}{1-\alpha m}\lambda_{N-1}^3 + (k^2\alpha x_{N-1}^2 - \frac{k}{2}x_{N-1}^2)\lambda_{N-1}^2] - \frac{m}{2}(x_N - x_{N-1})^2 = 0 $$ I am required to factorize it to find the value of $\lambda_{N-1}$. I would normally use long division to find cubic roots. But in this case, using that would be too complicated. Is there another way to approach this?

Answer 1

I'm not sure if this is a right way to solving this, but I managed to factorize it to solve it. $$ [\frac{k^3\alpha^2x_{N-1}^2}{1-\alpha m}\lambda_{N-1}^3 + (k^2\alpha x_{N-1}^2 - \frac{k}{2}x_{N-1}^2)\lambda_{N-1}^2] - \frac{m}{2}(x_N - x_{N-1})^2 = 0 $$ $$ [(kx_{N-1}^2)\lambda_{N-1}^2](\frac{k^2\alpha^2}{1-\alpha m}\lambda_{N-1} + k\alpha -\frac{1}{2}) = \frac{m}{2}(x_N - x_{N-1})^2 $$ $$ \lambda_{N-1}^2 = \frac{m}{2k}\frac{(x_N - x_{N-1})^2}{x_{N-1}^2}\\ or \\ (\frac{k^2\alpha^2}{1-\alpha m}\lambda_{N-1} + k\alpha -\frac{1}{2}) = \frac{m}{2}(x_N - x_{N-1})^2 $$ from where i can solve this for $\lambda_{N-1}$