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A prime number expressible as $(x^3-y^3)/(x-y)=x^2+xy+y^2$ for specified consecutive or nearly consecutive positive integers. The first kind has $x=y+1$ and therefore the form $3y^2+3y+1$; the second kind has $x=y+2$ and therefore the form $3y^2+6y+4$. The first kind arises when a difference of two cubes is itself prime: the factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$ forces $x-y=1$ for $x>y>0$.

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.