$3 x^{3}-2 x^{2}+1=0$ সমীকরণের মূলগুলো $\alpha, \beta$ এবং $\gamma$ হলে $\sum \alpha^2 \beta=?$

$3 x^{3}-2 x^{2}+1$ $=3 x^{3}-2 x^{2}$ $+0 x+1=0$ $\therefore$ মূলগুলি, $\alpha, \beta, \gamma$ $\therefore \alpha+\beta+\gamma$ $=\frac{2}{3}, \alpha \beta \gamma$ $=-\frac{1}{3}, \alpha$ $\beta+\beta \gamma+\gamma$ $\alpha=0$ এখন, $\sum \alpha^{2} \beta$ $=\alpha^{2} \beta+\alpha^{2} \gamma$ $+\beta^{2} \gamma+\beta^{2}$ $\alpha+\gamma^{2} \alpha$ $+\gamma^{2} \beta$ $=\left(\alpha^{2} \beta+\alpha \beta^{2}+\alpha \beta \gamma\right)$ $+\left(\alpha \beta \dot{\gamma}+\beta^{2} \gamma+\beta \gamma^{2}\right)$ $+\left(\alpha^{2} \gamma+\alpha \beta \gamma+\alpha \gamma^{2}\right)-$ $3 \alpha \beta \gamma$ $=\alpha \beta(\alpha+\beta+\gamma)$ $+\beta \gamma(\alpha+\beta+\gamma)$ $+\gamma \alpha(\alpha+\beta+\gamma)-$ $3 \alpha \beta \gamma$ $=(\alpha+\beta+\gamma)$ $(\alpha \beta+\beta \gamma+\gamma \alpha)$ $-3 \alpha \beta \gamma$ $=\frac{2}{3} \cdot 0-$ $3\left(\frac{-1}{3}\right)=1$