If $a^2+b^2-c^2=0$, then the value of $\frac{2\left(a^6+b^6-c^6\right)}{3 a^2 b^2 c^2}$ is:

-2

If $a^2+b^2-c^2=0$

Then the value of $\frac{2\left(a^6+b^6-c^6\right)}{3 a^2 b^2 c^2}$ = ?

Let the values of a = 1 , b = 1 and c = \(\sqrt {2}\) . These values will satisfy the equation.

Put these values in the required equation = $\frac{2\left(1^6+1^6- \sqrt {2} )^6\right)}{3 (1)^2 (1)^2 \sqrt {2} )^2}$

= 2 × \(\frac{-6}{6}\) = -2