সৃজনশীলঃ ‘দৃশ্যকল্প-১: mx2 +…’ চট্টগ্রাম বোর্ড ২০২২ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, ${\mathrm{mx}}^2+\mathrm{nx}+p=0............\left(1\right)$

${\mathrm{px}}^2-4\mathrm{nx}+16m=0...........\left(2\right)$

(1) নং সমীকরণের মূলদ্বয় $\alpha$ ও হলে,

$\alpha +\beta =-\frac{n}{m}$ এবং $\alpha \beta =\frac{p}{m}$

এখন, সমীকরণ (2) কে m দ্বারা ভাগ করে পাই,

$\frac{p}{m}{x}^2-\frac{4n}{m}x+16=0$

বা, $\frac{p}{m}{x}^2+4(-\frac{n}{m})x+16=0$

বা, $\alpha \beta x^2+4(\alpha +\beta )x+16=0[\frac{p}{m}$ এবং $-\frac{n}{m}$ এর মান বসিয়ে ]

বা, $\alpha \beta x^2+4\alpha x+4\beta x+16=0$

বা, $\alpha x(\beta x+4)+4(\beta x+4)=0$

বা, $(\beta x+4)(\alpha x+4)=0$

বা, $\beta x+4=0$

অথবা, $\alpha x+4=0$

বা, $\alpha x=-4$

$\therefore x=-\frac{4}{\alpha }$

বা, $\beta x=-4$

$\therefore x=-\frac{4}{\beta }$

অর্থাৎ সমীকরণ (2) এর মূলদ্বয় $\frac{-4}{\alpha }$ এবং $\frac{-4}{\beta }$.

এখানে, $x^3+dx+h=0$

সমীকরণটির মূলত্রয় $\alpha ,\beta ,\gamma$ হলে,

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

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{d}{1}=d$ এবং $\alpha \beta \gamma =-\frac{h}{1}=-h$

প্রদত্ত রাশি, $=\sum \frac{1}{{\alpha }^3}=\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}+\frac{1}{{\gamma }^3}$

$=\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}+\frac{1}{{\gamma }^3}-\frac{3}{\alpha \beta \gamma }+\frac{3}{\alpha \beta \gamma }$

$=(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}-\frac{1}{\alpha \beta }-\frac{1}{\beta \gamma }-\frac{1}{\gamma \dot{\alpha }})+\frac{3}{\alpha \beta \gamma }$

$[\because a^3+b^3+c^3-3abc=(a+b+c\left)(a^2+b^2+c^2-ab-bc-ca)\right]$

$=\left(\frac{\beta \gamma +\gamma \alpha +\alpha \beta }{a\beta y}\right)\{{(\frac{1}{a}+\frac{1}{\beta }+\frac{1}{y})}^2-2(\frac{1}{a\beta }+\frac{1}{\beta y}+\frac{1}{\gamma a})-(\frac{1}{a{\beta }^2}+\frac{1}{\beta y}+\frac{1}{\gamma \alpha })\}+\frac{3}{a\beta y}$

$[\because a^2+b^2+c^2=(a+b+c{)}^2-2(ab+bc+ca)]$

$=\left(\frac{\alpha \gamma +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)\{{\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)}^2-3(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })\}+\frac{3}{\alpha \beta \gamma }$

$=\left(\frac{d}{-h}\right)\{{\left(\frac{d}{-h}\right)}^2-3\left(\frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }\right)\}+\frac{3}{-h}$

$=\left(\frac{d}{-h}\right)\{\frac{d^2}{h^2}-3\cdot \frac{0}{h}\}-\frac{3}{h}$

$=\frac{-d}{h}\times \left(\frac{d^2}{h^2}\right)-\frac{3}{h}=-\frac{d^3}{h^3}-\frac{3}{h}=\frac{-(d^3+3h^2)}{h^3}$

নির্ণেয় মান : $\frac{-(d^3+3h^2)}{h^3}$.