When does $x^3+y^3=kz^2$? For which integers $k$ does
$$
x^3+y^3=kz^2
$$
have a solution with $z\ne0$ and $\gcd(x,y)=1$? Is there a technique for counting the number of solutions for a given $k$?

Consider the equation: $$(a+b)^3+(a-b)^3=2a(a^2+3b^2)$$ Now let $a^2+3b^2=c^2$. We want to find a general equation for this. Note that $$(m-n)^2+2mn=(m+n)^2$$ Therefore, substitute $m\mapsto m^2$ and $n\mapsto n^2k$ then $$(m^2-k\cdot n^2)^2+k(2mn)^2=(m^2+k\cdot n^2)^2$$ Let $k=3$, and it follows that $a=m^2-3n^2$ and $b=2mn$, given that $a^2+b^2=(m^2+3n^2)^2$. $$\therefore (m^2-3n^2+2mn)^3+(m^2-3n^2-2mn)^3=2(m^2-3n^2)(m^2+3n^2)^2$$ We could also substitute  $m\to 3m^2$ and $n\mapsto n^2$ to get $$(3m^2-n^2+2mn)^3+(3m^2-n^2-2mn)^3=2(3m^2-n^2)(3m^2+n^2)^2$$
