Given, $\cos \alpha+3 \cos 3 \beta+5 \cos 5 \gamma=0$ ...(i)
$\sin \alpha+3 \sin 3 \beta+5 \sin 5 \gamma=0$ ...(ii)
$\cos 3 \alpha+27 \cos 9 \beta+125 \cos 15 \gamma$
$=\left(\lambda^2-4\right) \cos (\alpha+3 \beta+5 \gamma)$ ...(iii)
To find, $\lambda=$ ?
Let $a=\cos \alpha+i \sin \alpha$
$\begin{aligned} & b=3(\cos 3 \beta+i \sin 3 \beta) \\ & c=5(\cos 5 \gamma+i \sin 5 \gamma)\end{aligned}$
Consider, $a+b+c$
$\begin{aligned}=(\cos \alpha+i \sin \alpha)+ & 3(\cos 3 \beta+i \sin 3 \beta) \\ & +5(\cos 5 \gamma+i \sin 5 \gamma)\end{aligned}$
$\begin{aligned}=(\cos \alpha+3 & \cos 3 \beta+5 \cos 5 \gamma) \\ & +i(\sin \alpha+3 \sin 3 \beta+5 \sin 5 \gamma)\end{aligned}$
$=0$
$\because \quad a^3+b^3+c^3=3 a b c$, if $a+b+c=0$
$\begin{aligned}(\cos \alpha+i \sin \alpha)^3+\{3(\cos 3 \beta+ & i \sin 3 \beta)\}^3 \\ & +\{5(\cos 5 \gamma+i \sin 5 \gamma)\}^3\end{aligned}$
$\begin{array}{r}=3(\cos \alpha+i \sin \alpha)\{3(\cos 3 \beta+i \sin 3 \beta)\} \\ \quad\{5(\cos 5 \gamma+i \sin 5 \gamma)\}\end{array}$
$\begin{aligned}=[\cos 3 \alpha+i \sin 3 \alpha)+ & \{27(\cos 9 \beta+i \sin 9 \beta)\} \\ & +\{25(\cos 15 \gamma+i \sin 15 \gamma)\}\end{aligned}$
$\begin{array}{r}=45(\cos \alpha+i \sin \alpha)(\cos 3 \beta+i \sin 3 \beta) \\ \quad(\cos 5 \gamma+i \sin 5 \gamma)\end{array}$
$\begin{array}{r}(\cos 3 \alpha+27 \cos 9 \beta+125 \cos 15 \gamma) \\ =45 \cos (\alpha+3 \beta+5 \gamma)\end{array}$
$\begin{aligned} & \therefore & \lambda^2-4 & =45 \\ \Rightarrow & & \lambda^2 & =49 \\ \Rightarrow & & \lambda & = \pm 7\end{aligned}$