সৃজনশীলঃ ‘x3+px+q=0 এর মুলগুলি…’ রাজশাহী কলেজ ২০২৫ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $x^3+\mathrm{px}+q=0$ সমীকরণের মূলত্রয় $\alpha ,\beta ,\gamma$ 

$\therefore \alpha +\beta +\gamma =0,\alpha \beta +\beta \gamma +\gamma \alpha =p,\alpha \beta \gamma =-q$ 

প্রদত্ত রাশি $=\frac{{\alpha }^3}{\beta +\gamma }+\frac{{\beta }^3}{\gamma +\alpha }+\frac{{\gamma }^3}{\alpha +\beta }$ 

$\begin{array}{cc}& =\frac{{\alpha }^3}{(\alpha +\beta +\gamma )-\alpha }+\frac{{\beta }^3}{(\alpha +\beta +\gamma )-\beta }+\frac{{\gamma }^3}{(\alpha +\beta +\gamma )-\gamma }\\ & =\frac{{\alpha }^3}{0-\alpha }+\frac{{\beta }^3}{0-\beta }+\frac{{\gamma }^3}{0-\gamma }\\ & =-{\alpha }^2-{\beta }^2-{\gamma }^2\end{array}$ 

$\begin{array}{cc}& =-({\alpha }^2+{\beta }^2+{\gamma }^2)\\ & =-\left\{\right(\alpha +\beta +\gamma ){\}}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha \left)\right\}\\ & =-(0-2p)\\ & =2p\end{array}$ 

নির্ণেয় মান = 2p. 

দেওয়া আছে , $x^3+\mathrm{px}+q=0$ সমীকরণেন মূনগুলি $\alpha ,\beta ,\gamma$ 

$\therefore \alpha +\beta +\gamma =0,\alpha \beta +\beta \gamma +\gamma \alpha =p,\alpha \beta \gamma =-q$ 

নির্ণেয় সমীকরনের মুলগুলি,

$\frac{\alpha +\beta }{{\gamma }^2}$ ,$\frac{\beta +\gamma }{{\alpha }^2},\frac{\gamma +\alpha }{{\beta }^2}$  

বা, $\frac{(\alpha +\beta +\gamma )-\gamma }{{\gamma }^2},\frac{(\alpha +\beta +\gamma )-\alpha }{{\alpha }^2},\frac{(\alpha +\beta +\gamma )-\beta }{{\beta }^2}$

বা, $\frac{0-\gamma }{{\gamma }^2},\frac{0-\alpha }{{\alpha }^2},\frac{0-\beta }{{\beta }^2}$

বা, $-\frac{1}{\gamma },-\frac{1}{\alpha },-\frac{1}{\beta }$ 

$\therefore -\frac{1}{\alpha }-\frac{1}{\beta }-\frac{1}{\gamma }=-(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })=-\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)=-\frac{p}{-q}=\frac{p}{q}$ 

আবার, $(-\frac{1}{\alpha })(-\frac{1}{\beta })+(-\frac{1}{\beta })(-\frac{1}{\gamma })+(-\frac{1}{\gamma })(-\frac{1}{\alpha })$ 

$=\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha }=\frac{\gamma +\alpha +\beta }{\alpha \beta \gamma }=\frac{0}{-q}=0$ 

এবং $(-\frac{1}{\alpha })(-\frac{1}{\beta })(-\frac{1}{\gamma })=-\frac{1}{\alpha \beta \gamma }=-\frac{1}{-q}=\frac{1}{q}$

নির্ণেয় সমীকরণ, $x^3-\left(\frac{p}{q}\right)x^2+\left(0\right)x-\left(\frac{1}{q}\right)=0$

বা, $x^3-\frac{p}{q}{x}^2-\frac{1}{q}=0$

$\therefore q{x}^3-p{x}^2-1=0$

নির্ণেয় সমীকরণ, ${\mathrm{qx}}^3-{\mathrm{px}}^2-1=0$.