Diophantine Equation: $a^3=a(b^2+c^2+d^2)+2bcd$ Let $a,b,c,d\in \mathbb{Z}$. Solve $$a^3=a(b^2+c^2+d^2)+2bcd$$
I've tried everything but I haven't been able to find a general solution.
Note: We may assume $\gcd(a,b,c,d)=1$ because of homogeneity.

Below mentioned equation from above has another parametrization, but without the use of a pythagorean triple,
$a^3=a(b^2+c^2+d^2)+2bcd$
$a=(k^2-4k+4)$
$b=(2k^2-4k)$
$c=-(7k^2+4k-4)$
$d=(2k^2-4k)$
