Prove that: `|alphabetagammaalpha^2beta^2gamma^2beta+gammagamma+alphaalpha+beta|=(alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma)` .

Answer

Proved

Explanation

$$\operatorname{LHS} \Delta=\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta\end{array}\right|$$
$$\Delta=\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \alpha+\beta+\gamma & \alpha+\beta+\gamma & \alpha+\beta+\gamma\end{array}\right| \quad\left\}\right.$$ Applying $$\left.R_{3} \rightarrow R_{1}+R_{3}\right\}$$
$$=(\alpha+\beta+\gamma)\left|\begin{array}{ccc}\alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ 1 & 1 & 1\end{array}\right|$$
$$=(\alpha+\beta+\gamma)\left|\begin{array}{ccc}\alpha & \beta-\alpha & \gamma-\alpha \\ \alpha^{2} & \beta^{2}-\alpha^{2} & \gamma^{2}-\alpha^{2} \\ 1 & 0 & 0\end{array}\right| \quad\left\{\right.$$ Applying $$\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1}, \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1}$$.
$$=(\alpha+\beta+\gamma)(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc}\alpha & 1 & 1 \\ \alpha^{2} & \beta+\alpha & \gamma+\alpha \\ 1 & 0 & 0\end{array}\right|$$
Taking out `(\beta-\alpha)` and `(\gamma-\alpha)` from `C_{2}` and `C_{3}` respectively.
$$
\begin{aligned}
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