In a ∆ABC, if
Question:

In a ∆ABC, if a4 + b4 + c4 = 2a2b2 + 2b2c2, then B = __________.

Solution:

In a ∆ABC,

Given $a^{4}+b^{4}+c^{4}=2 a^{2} b^{2}+2 b^{2} c^{2}$

i.e $a^{4}+b^{4}-2 a^{2} b^{2}+c^{2}-2 b^{2} c^{2}=0$

i.e $\left(a^{2}+c^{2}\right)^{2}+b^{4}-2 a^{2} b^{2}-2 b^{2} c^{2}-2 a^{2} c^{2}=0$

i.e $\left(\underline{a}^{2}+c^{2}\right)^{2}+b^{4}-2 b^{2}\left(\underline{a}^{2}+c^{2}\right)-2 a^{2} c^{2}=0$

i.e $\left(a^{2}+c^{2}-b^{2}\right)^{2}=2 a^{2} c^{2}$

i.e $a^{2}+c^{2}-b^{2}=\sqrt{2} a c$

Using cosine formula,

i.e $\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{\sqrt{2} a c}{2 a c}$

$\cos B=\frac{1}{\sqrt{2}}$

i.e $B=\frac{\pi}{4}$ or $\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$