সৃজনশীলঃ ‘উদ্দীপক-১: ax3+bx+c=0 সমীকরণের…’ বরিশাল বোর্ড ২০২২ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

${\mathrm{ax}}^3+\mathrm{bx}+c=0$ সমীকরণের মূলত্রয় $\alpha ;\beta ,\gamma$.

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

$\therefore \alpha \beta +\beta \gamma +\gamma \alpha =\frac{b}{a}$

$\therefore \alpha \beta \gamma =-\frac{c}{a}$

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

$x^3-(\frac{{\gamma }^2}{\alpha +\beta }+\frac{{\alpha }^2}{\beta +\gamma }+\frac{{\beta }^2}{\gamma +\alpha })x^2+(\frac{{\gamma }^2}{\alpha +\beta }\cdot \frac{{\alpha }^2}{\beta +\gamma }+\frac{{\alpha }^2}{\beta +\gamma }\cdot \frac{{\beta }^2}{\gamma +\alpha }+\frac{{\gamma }^2}{\alpha +\beta }\cdot \frac{{\beta }^2}{\gamma +\alpha })x-\frac{{\gamma }^2}{\alpha +\beta }\cdot \frac{{\alpha }^2}{\beta +\gamma }\cdot \frac{{\beta }^2}{\gamma +\alpha }=0$

বা, $x^3-(\frac{{\gamma }^2}{-\gamma }+\frac{{\alpha }^2}{-\alpha }+\frac{{\beta }^2}{-\beta })x^2+(\frac{{\gamma }^2{\alpha }^2}{(-\gamma )(-\alpha )}+\frac{{\alpha }^2\cdot {\beta }^2}{(-\alpha )(-\beta )}+\frac{{\gamma }^2\cdot {\beta }^2}{(-\gamma )(-\beta )})x-(\frac{{\gamma }^2}{-\gamma },\frac{{\alpha }^2}{-\alpha }\cdot \frac{{\beta }^2}{-\beta })=0$

বা, $x^3+(\gamma +\alpha +\beta )x^2+(\gamma \alpha +\alpha \beta +\beta \gamma )x+\gamma \alpha \beta =0$

বা, $x^3+(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\gamma \alpha )x+\alpha \beta \gamma =0$

বা, $x^3+0\cdot x^2+\frac{b}{a}\cdot x-\frac{c}{a}=0$

$\therefore {\mathrm{ax}}^3+\mathrm{bx}-c=0$

নির্ণেয় সমীকরণ : ${\mathrm{ax}}^3+\mathrm{bx}-c=0$.

উদ্দীপক-২ হতে পাই.

$(y+ir{)}^{\frac{1}{3}}=(a+ib)$

বা, $(y+ix)=(a+ib{)}^3$ [উভয় পক্ষকে ঘন করে]

বা, $(y+ix)=a^3+3a^2\cdot ib+3a(ib{)}^2+(ib{)}^3$

বা, $(y+ix)=x^3+3a^{2}bi+3a{b}^2(-1)+i^3{b}^3[\because i^2=-1]$

বা, $(y+ix)=a^3-3a{b}^2+3a^{2}bi-i{b}^3$

বা, $(y+ix)=(a^3-3a{b}^2)+i(3a^{2}b-b^3)$ ......... (i)

(i) নং এর উভয়পক্ষের বাস্তব ও কাল্পনিক অংশ সমীকৃত করে পাই,

$y=a^3-3a{b}^2$ এবং $x=3a^{2}b-b^3$

বামপক্ষ = ax + by

$\begin{array}{cc}& =a(3a^{2}b-b^3)+b(a^3-3a{b}^2)\\ & =ab(3a^2-b^2)+ab(a^2-3b^2)\\ & =ab(3a^2-b^2+a^2-3b^2)\\ & =ab(4a^2-4b^2)=4ab(a^2-b^2)\end{array}$ = ডানপক্ষ

$\therefore \mathrm{ax}+\mathrm{by}=4\mathrm{ab}(a^2-b^2)\cdot$ (দেখানো হলো)