Demonstrate using determinant properties that the determinant of $A$ is equal to $2abc(a+b+c)^3$

Solution 1:

the value of the determinant is evidently a homogeneous symmetric polynomial D(a,b,c) of degree six.

it is easy to see that it vanishes if $a=0$, $b=0$ or $c=0$.

and if $a+b+c=0$ then $(b+c)^2=a^2$ etc. , so again the determinant vanishes.

hence for suitably chosen $\lambda$ and $\mu$ $$ D(a,b,c) = abc(a+b+c)\left(\lambda (a^2+b^2+c^2) + \mu(ab+bc+ca) \right) $$ we may easily compute $D(1,1,1)=54$, hence $$ \lambda+\mu = 6 $$ likewise $D(2,1,1) = 256$, giving: $$ 6\lambda +5 \mu = 32 $$thus $\lambda=2$ and $\mu=4$, so $$ \lambda (a^2+b^2+c^2) + \mu(ab+bc+ca) = 2(a+b+c)^2 $$ and, finally: $$ D(a,b,c) = 2abc(a+b+c)^3 $$