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The difference of two cubes identity gives the real polynomial factorizationExpanding the right side cancels the two mixed terms of each degree. The quadratic factor satisfies $a^2+ab+b^2=(a+b/2{)}^2+3b^2/4$.

Replacing $b$ by $-b$ in the difference of two cubes givesFor real cube roots $u=a^{1/3}$ and $v=b^{1/3}$, this simplifies $(a^{1/3}+b^{1/3})(a^{2/3}-(ab{)}^{1/3}+b^{2/3})$ to $a+b$. Specifying real cube roots matters when using such identities with negative real inputs; unconstrained complex exponentiation has logarithm-branch choices.