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Question Number 195998 by Frix last updated on 15/Aug/23

We can transform to get rid of the \sqrt[3]{\dots}

a^{\frac{1}{3}} + b^{\frac{1}{3}} = c^{\frac{1}{3}}

{(a^{\frac{1}{3}} + b^{\frac{1}{3}})}^{3} = c

a + b + 3 a^{\frac{1}{3}} b^{\frac{1}{3}} (a^{\frac{1}{3}} + b^{\frac{1}{3}}) = c

3 a^{\frac{1}{3}} b^{\frac{1}{3}} c^{\frac{1}{3}} = c - a - b

27 a b c = {(c - a - b)}^{3}

Is it possible to do the same for

a^{\frac{1}{3}} + b^{\frac{1}{3}} = c^{\frac{1}{3}} + d^{\frac{1}{3}}

I found no path yet \dots