How to approach this equation? $(1+2x+3x^{2})^{6}=0$
Sorry, I am not even sure how to approach this. I though of simplifying it first by assuming $1$, $x$ and $x^{2}$ as cube roots of 1 (complex numbers). However, I am not sure if I am at liberty to assume that. And even then it seems that I still have to expand it.
You have $$(1+2x+3x^2)^6=0$$ if $$1+2x+3x^2=0$$ since the exponent isn't zero.
$%edits$ By the quadratic formula we get $$x^2+\frac{2}{3}x+\frac{1}{3}=0$$ so $$x_{1,2}=-\frac{1}{3}\pm\sqrt{\frac{1}{9}-\frac{1}{3}}$$ so no real solutions.