Solving equation of degree 5

Question

Find all real values of $x$ such that $$x^2+1=2\sqrt[3]{2x-1}.$$ Let $t=\sqrt[3]{2x-1}$. Then the equation is equivalent to $$\left(\frac{t^3+1}{2}\right)^2+1-2t=0 \Leftrightarrow t^6+2t^3-8t+5=0 \Leftrightarrow \left(t-1\right)\left(t^5+t^4+t^3+3t^2+3t-5\right)=0$$ I cannot handle the equation of degree $5$. Please help me. Thank you very much.

Answer 1

Taking everything to the third power gives $$(x^2+1)^3=2^3(2x-1)$$ that is $$x^6+3x^4+3x^2+1=16x-8,$$ i.e. we are searching for roots of $$x^6+3x^4+3x^2-16x+9=0.$$ Since the sum over all of the coefficients is zero, $x=1$ is a root. Polynomial division gives that the other solutions are the roots of $$x^5+x^4+4x^3+4x^2+7x-9.$$ You get four complex roots and another real root that you can calculate using a computer algebra system such was WolframAlpha.