Identity cosines

Relevant wiki: Proving Trigonometric Identites

$\begin{aligned} \alpha{\cos}^2 3\theta + \beta{\cos}^4\theta & = 16 {\cos}^6\theta + 9 {\cos}^2\theta\\ \\ \alpha{\left(4\cos^3\theta - 3\cos\theta\right)}^2 + \beta{\cos}^4\theta & = 16 \cos^6\theta + 9 \cos^2\theta\\ \\ \alpha\left(16 \cos^6\theta + 9 \cos^2\theta - 24 \cos^4\theta\right) + \beta \cos^4\theta & = 16 \cos^6\theta + 9 \cos^2\theta\\ \\ 16\alpha \cos^6\theta + 9\alpha \cos^2\theta - 24\alpha \cos^4\theta + \beta \cos^4\theta & = 16 \cos^6\theta + 9 \cos^2\theta\\ \\ 16\alpha \cos^6\theta + \left(\beta - 24\alpha\right)\cos^4\theta + 9\alpha \cos^2\theta & = 16 \cos^6\theta + 0 \cos^4\theta + 9 \cos^2\theta\\ \\ \text{Comparing the corresponding coefficients}:-\\ \\ 16\alpha & = 16\\ \alpha & = 1\\ \\ \beta - 24 \alpha & = 0\\ \beta - 24×1 & = 0\\ \beta & = 24\\ \\ \implies \alpha + \beta & = 1 + 24\\ & = \color{#3D99F6}{\boxed{25}} \end{aligned}$

We are told that $\alpha \cos^2 3 \theta + \beta \cos^4 \theta = 16 \cos^6 \theta + 9 \cos^2 \theta$ is an identity, and we are asked to find $\alpha + \beta$ . We can set $\theta = 0$ , to get $\alpha + \beta = 25$ .