A radical equation $(2x+1)^{2/3}+(2x-1)^{2/3}-2x^{2/3}=2^{1/3}$

Question

Solve the equation $(2x+1)^{2/3}+(2x-1)^{2/3}-2x^{2/3}=2^{1/3}$. I am looking for real roots. The graph of the equation tell us there are 4 solutions: roughly at $\pm0.09, \pm 1.64$, but I want to know if the roots can be found through algebraic manipulations. I tried letting $u=(2x+1)^{1/3}, v = (2x-1)^{1/3}$ and then reduce the equation to an expression with integer exponent, but the result doesn't seem to factorize easily. My guess is that there is some clever transformation which turns this equation to a quadratic or a cubic equation. Can anyone help?

Answer 1

Write it as:

$$ \sqrt[3]{(2x+1)^2} + \sqrt[3]{(2x-1)^2} = 2\sqrt[3]{x^2} + \sqrt[3]{2} $$

Cube both sides using $(a+b)^3=a^3+b^3+3ab(a+b)$:

$$ \require{cancel} (\cancel{4x^2}+\xcancel{4x}+\cancel{1}) + (\cancel{4x^2}-\xcancel{4x}+\cancel{1}) + 3\sqrt[3]{(2x+1)^2(2x-1)^2}\bcancel{\left(\sqrt[3]{(2x+1)^2} + \sqrt[3]{(2x-1)^2}\right)} \\ = \cancel{8x^2} + \cancel{2} + 6\sqrt[3]{2x^2}\bcancel{\left(2\sqrt[3]{x^2} + 2^{1/3}\right)} $$

Cube again:

$$ \begin{align} 27(4x^2-1)^2 &= 432 x^2 \\ 16 x^4 - 24 x^2 + 1 &= 0 \end{align} $$

The latter is a biquadratic in $x$ with $4$ real roots.

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[ EDIT ] $\;$ Verification using WA's GroebnerBasis here.

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[ EDIT #2 ] $\;$ Canceling the factors in parentheses in the second step assumes that they are not zero. It is straightforward to verify that the simultaneous equations $\sqrt[3]{(2x+1)^2} + \sqrt[3]{(2x-1)^2}$ $= 2\sqrt[3]{x^2} + \sqrt[3]{2} = 0$ have no real root, so the cancelation is justified. This, however, explains why the WA solution includes the complex roots $x = \pm \frac{i}{2}\,$.