Question
( a^{4}+b^{4}+c^{4}+a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}=2 a^{2} c^{2}+2 c^{2} b^{2} )
( a^{4}+b^{4}+c^{2}+a^{2} b^{2}-a^{2} c^{2}-b^{2} c^{2}=a^{2} c^{2}+a^{2} b^{2}+a^{2} b^{2} )
( left(a^{2}+b^{2}-c^{2}right)^{2}=a^{2} c^{2}+a^{2} b^{2} )
( cos (c)=frac{a^{2}+b^{2}-c^{2}}{2 a b}=pm sqrt{a^{2} c^{2}+c^{2} b^{2}}=frac{c^{2}}{2 a b} )
( =frac{pm a^{*}(sqrt{e^{2}+b^{2}})}{2 sqrt{b}}=left(pm frac{1}{2}right) )
fiomabous
( begin{aligned} &=pm sqrt{frac{b^{2}+c^{2}}{2 b}} Rightarrow & c=60^{circ} text { or } frac{12^{0}}{1} end{aligned} )

3. If a4 + b4 + c4 + a2b2 = 2c2(a2 + b2) then show that, C = 60° or 120°
Solution
