Solutions Of Triangles

$a^4+b^4+c^4=2c^2a^2+2c^2b^2$

$\implies a^4+b^4+c^4-2c^2a^2-2c^2b^2+2a^2b^2=2a^2b^2$

$\implies (a^2+b^2-c^2)^2=2a^2b^2$

$\implies a^2+b^2-c^2=\sqrt{2}ab$

$\implies \dfrac{a^2+b^2-c^2}{ab} =\sqrt{2}$

$\implies \dfrac{a^2+b^2-c^2}{2ab} = \dfrac{1}{\sqrt{2}}$

BY COSINE RULE:

$\implies \cos C = \dfrac{1}{\sqrt{2}}$

$\implies C=45^{o}$

Also for $C=135$ we missed out this portion that if

$\implies (a^2+b^2-c^2)^2=2a^2b^2$

then it can also be written

$\implies (c^2-b^2-a^2)^2=2a^2b^2$

$\implies c^2-b^2-a^2=\sqrt{2}ab$

So again by same steps we get $C=135^{o}$