(ক) আমরা জানি, $\cos 3A = \cos(2A + A) = \cos 2A \cos A - \sin 2A \sin A$। $$ \begin{aligned} \cos 3A &= (2\cos^2 A - 1)\cos A - (2\sin A \cos A)\sin A \\ &= 2\cos^3 A - \cos A - 2\cos A \sin^2 A \\ &= 2\cos^3 A - \cos A - 2\cos A(1 - \cos^2 A) \\ &= 2\cos^3 A - \cos A - 2\cos A + 2\cos^3 A \\ &= 4\cos^3 A - 3\cos A \end{aligned} $$ (খ) ত্রিভুজ $ABC$ এর ক্ষেত্রে, $A+B+C = \pi$ বা $A = \pi - (B+C)$। $$ \begin{aligned} \text{বামপক্ষ} &= \cos(B-C) + \cos A \\ &= \cos(B-C) + \cos(\pi - (B+C)) \\ &= \cos(B-C) - \cos(B+C) \\ &= 2 \sin B \sin C \end{aligned} $$ সাইন সূত্রানুসারে, $b = 2R \sin B$ এবং $c = 2R \sin C$, অর্থাৎ $\sin B = \frac{b}{2R}$ এবং $\sin C = \frac{c}{2R}$। $$ \begin{aligned} \text{বামপক্ষ} &= 2 \left( \frac{b}{2R} \right) \left( \frac{c}{2R} \right) \\ &= \frac{bc}{2R^2} \\ &= \text{ডানপক্ষ} \end{aligned} $$ প্রমাণিত। (গ) দেওয়া আছে, $a^4 + b^4 + c^4 = 2c^2(a^2 + b^2)$। $$ \begin{aligned} \Rightarrow\quad a^4 + b^4 + c^4 - 2c^2a^2 - 2c^2b^2 &= 0 \\ \Rightarrow\quad a^4 + b^4 + c^4 - 2c^2a^2 - 2c^2b^2 + 2a^2b^2 &= 2a^2b^2 \\ \Rightarrow\quad (a^2 + b^2 - c^2)^2 &= 2a^2b^2 \\ \Rightarrow\quad a^2 + b^2 - c^2 &= \pm \sqrt{2}ab \\ \Rightarrow\quad \frac{a^2 + b^2 - c^2}{2ab} &= \pm \frac{\sqrt{2}ab}{2ab} \\ \Rightarrow\quad \cos C &= \pm \frac{1}{\sqrt{2}} \end{aligned} $$ যদি $\cos C = \frac{1}{\sqrt{2}} = \cos 45^\circ$, তবে $C = 45^\circ$। যদি $\cos C = -\frac{1}{\sqrt{2}} = -\cos 45^\circ = \cos(180^\circ - 45^\circ) = \cos 135^\circ$, তবে $C = 135^\circ$। সুতরাং, $C = 45^\circ$ অথবা $135^\circ$। দেখানো হলো।