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.
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There is an error in your computation. A quick check using WolframAlpha gave https://www.wolframalpha.com/input/?i=roots+t%5E5%2Bt%5E4%2Bt%5E3%2B3t%5E2%2B3t-5 as solutions for the polynomial whereas the original equation https://www.wolframalpha.com/input/?i=solve+x%5E2%2B1%3D2*(2x-1)%5E(1%2F3)+for+x admits only two solutions. – Feb 13 '19 at 13:19
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This equation is from AOPS? – Dr. Sonnhard Graubner Feb 13 '19 at 13:19
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Note $x=1$ is a solution – J. W. Tanner Feb 13 '19 at 13:21
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We all know this from above! – Dr. Sonnhard Graubner Feb 13 '19 at 13:27
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It has 2 root, 1 and 0.716. But I a want an explicit formula of the root 0.7168 – RuaSun Feb 13 '19 at 13:38
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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.
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You can but in general won't have an explicit formula for the roots of a degree five polynomial. That's the statement of the Abel-Ruffini theorem. For a discussion see here https://math.stackexchange.com/questions/1555743/how-do-you-solve-5th-degree-polynomials – Feb 13 '19 at 13:34
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The decimal approximation for the remaining real root is 0.71683662408618024174708229 if that helps you. – Feb 13 '19 at 13:35