Question

If in the $∆ABC$, $\angle B=45°$, then $a^4+b^4+c^4$ is equal to:

• A. $2a^2\left(b^2+c^2\right)$
• B. $2c^2\left(a^2+b^2\right)$
• C. $2b^2\left(a^2+c^2\right)$
• D. $2\left(a^2{b}^2+b^2{c}^2+c^2{a}^2\right)$

Answer 1

The correct option is C

$2b^2\left(a^2+c^2\right)$

Explanation for the correct option:

Find the value of $a^4+b^4+c^4$:

It is given that $\angle B=45°$, by applying the cosine formula, we get

$\begin{array}{rcl}{b}^2& =& a^2+c^2-2ac\cos B\\ & \Rightarrow & 2ac\cos 45°=a^2+c^2-b^2\\ \Rightarrow 2ac\left(\frac{1}{\sqrt{2}}\right)& =& a^2+c^2-b^2\end{array}$

By squaring both sides, we get

$\begin{array}{rcl}\frac{4a^2{c}^2}{2}& =& {\left(a^2+c^2-b^2\right)}^2\\ \Rightarrow 2a^2{c}^2& =& a^4+b^4+c^4+2a^2{c}^2-2b^2{c}^2-2a^2{b}^2\\ & \Rightarrow & a^4+b^4+c^4=2b^2{c}^2+2a^2{b}^2\\ & \Rightarrow & a^4+b^4+c^4=2b^2\left(c^2+a^2\right)\end{array}$

Hence, option C is correct.