Let $P =\begin{bmatrix}\cos\frac{π}{9}&\sin\frac{π}{9}\\-\sin\frac{π}{9}&\cos\frac{π}{9}\end{bmatrix}$ and $α,β,γ$ be non-zero real numbers such that $αP^6+ βP^3 + γI$ is the zero matrix. Then, $(α^2+β^2+γ^2)^{(α-β) (β-γ) (γ-α)}$ is

1

We have,

$P =\begin{bmatrix}\cos\frac{π}{9}&\sin\frac{π}{9}\\-\sin\frac{π}{9}&\cos\frac{π}{9}\end{bmatrix}$

$∴P^2=PP=\begin{bmatrix}\cos\frac{π}{9}&\sin\frac{π}{9}\\-\sin\frac{π}{9}&\cos\frac{π}{9}\end{bmatrix}\begin{bmatrix}\cos\frac{π}{9}&\sin\frac{π}{9}\\-\sin\frac{π}{9}&\cos\frac{π}{9}\end{bmatrix}$

$⇒P^2=\begin{bmatrix}\cos\frac{2π}{9}&\sin\frac{2π}{9}\\-\sin\frac{2π}{9}&\cos\frac{2π}{9}\end{bmatrix}$

In general, $P^n=\begin{bmatrix}\cos\frac{nπ}{9}&\sin\frac{nπ}{9}\\-\sin\frac{nπ}{9}&\cos\frac{nπ}{9}\end{bmatrix},n∈N$

$∴P^3=\begin{bmatrix}\cos\frac{π}{3}&\sin\frac{π}{3}\\-\sin\frac{π}{3}&\cos\frac{π}{3}\end{bmatrix}=\begin{bmatrix}\frac{1}{2}&\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&\frac{1}{2}\end{bmatrix}$

and, $P^6=\begin{bmatrix}\cos\frac{2π}{3}&\sin\frac{2π}{3}\\-\sin\frac{2π}{3}&\cos\frac{2π}{3}\end{bmatrix}=\begin{bmatrix}-\frac{1}{2}&\frac{\sqrt{3}}{2}\\-\frac{\sqrt{3}}{2}&-\frac{1}{2}\end{bmatrix}$

It is given that $αP^6+ βP^3 + γI=O$

$⇒\begin{bmatrix}-\frac{α}{2}+\frac{β}{2}+γ&\frac{\sqrt{3}}{2}α+\frac{\sqrt{3}}{2}β\\-\frac{\sqrt{3}}{2}α-\frac{\sqrt{3}}{2}β&-\frac{α}{2}+\frac{β}{2}+γ\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$

$⇒-α+β+2γ=0,α+β=0,-α+β+2γ=0$

$⇒β=-α$ and $γ=α$

$∴(α^2+β^2+γ^2)^{(α-β) (β-γ) (γ-α)}=(α^2+β^2+γ^2)^0=1$