3x³-2x²+1=0 সমীকরণের মূলগুলো α, β, ɤ হলে, ∑α²β এর মান কত? 

$3 x^{3}-2 x^{2}+1=3 x^{3}-2 x^{2}+0 x+1=0$

$\therefore$ মূলগুলি $\alpha, \beta, \gamma$

$\begin{aligned} \therefore \alpha & +\beta+\gamma=\frac{2}{3}, \alpha \beta \gamma=-\frac{1}{3}, \alpha \beta+\beta \gamma+\gamma \alpha=0 \\ \text { এখন, } & \sum \alpha^{2} \beta=\alpha^{2} \beta+\alpha^{2} \gamma+\beta^{2} \gamma+\beta^{2} \alpha+\gamma^{2} \alpha+\gamma^{2} \beta \\ = & \left(\alpha^{2} \beta+\alpha \beta^{2}+\alpha \beta \gamma\right)+\left(\alpha \beta \gamma+\beta^{2} \gamma+\beta \gamma^{2}\right) \\ & +\left(\alpha^{2} \gamma+\alpha \beta \gamma+\alpha \gamma^{2}\right)-3 \alpha \beta \gamma \\ = & \alpha \beta(\alpha+\beta+\gamma)+\beta \gamma(\alpha+\beta+\gamma) \\ & +\gamma \alpha(\alpha+\beta+\gamma)-3 \alpha \beta \gamma \\ = & (\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha)-3 \alpha \beta \gamma \\ = & \frac{2}{3} \cdot 0-3\left(\frac{-1}{3}\right)=1 \end{aligned}$