I need an exact answer for $x$ in the following algebraic expression: $$ x^{4/3} - 2\sqrt{2}x^2 + 2\sqrt{2} = 0 $$ I don't need or want an approximate answer, I already know that $x≈1.20569$. So any exponential expressions (no matter the integer in the exponent is), cannot be evaluated; I need it (or them) to be left as an exponential expression(s). Any websites that I have used (including new websites that I've searched for) are not helpful, and do not give me an answer in the form that I need. (This is despite these websites working perfectly fine for less complicated equations). If you do manage to help me with finding $x$ in its exact form, then thank you very much in advance.
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$\begingroup$or alternatively we have $$x^{4/3}=2\sqrt{2}(x^2-1)$$ and after cube the equation we get $$x^4=(2\sqrt{2})^3(x^2-1)^3$$ this yields to $${x}^{4}-16\,\sqrt {2}{x}^{6}+48\,\sqrt {2}{x}^{4}-48\,\sqrt {2}{x}^{2} +16\,\sqrt {2} =0$$ which can factorize into $$1/2\,\sqrt {2} \left( -32\,{x}^{6}+\sqrt {2}{x}^{4}+96\,{x}^{4}-96\,{x }^{2}+32 \right) =0$$ note that $$x=1/24\,{\frac {\sqrt {6}\sqrt {\sqrt [3]{13826\,\sqrt {2}+576+768\, \sqrt {648+3\,\sqrt {2}}} \left( \left( 13826\,\sqrt {2}+576+768\, \sqrt {648+3\,\sqrt {2}} \right) ^{2/3}+\sqrt {2}\sqrt [3]{13826\, \sqrt {2}+576+768\,\sqrt {648+3\,\sqrt {2}}}+96\,\sqrt [3]{13826\, \sqrt {2}+576+768\,\sqrt {648+3\,\sqrt {2}}}+192\,\sqrt {2}+2 \right) }}{\sqrt [3]{13826\,\sqrt {2}+576+768\,\sqrt {648+3\,\sqrt {2}}}}} $$ is one solution of this equation, all Solutions can be expressed by radicals
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