সৃজনশীলঃ ‘দৃশ্যকল্প-১: x3+ax+b=0.দৃশ্যকল্প-২ :…’ সিলেট বোর্ড ২০২৫ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $x^3+\mathrm{ax}+b=0$
বা, $x^3+0\cdot x^2+ax+b=0$$.......\left(1\right)$ 

$\left(1\right)$ নং এর মূলত্রয় $\alpha ,\beta ,\gamma$ হলে, 

$\begin{array}{cc}& \alpha +\beta +\gamma =0\\ & \alpha \beta +\beta \gamma +\gamma \alpha =\frac{a}{1}=a\end{array}$ 

এবং $\alpha \beta \gamma =\frac{-\dot{b}}{1}=-b$ 

এখন, $\Sigma \frac{1}{{\alpha }^2}=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}$ 

$\begin{array}{cc}& ={(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma })}^2-2(\frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha })\\ & ={\left(\frac{\beta \gamma +\gamma \alpha +\alpha \beta }{\alpha \beta \gamma }\right)}^2-2\left(\frac{\gamma +\alpha +\beta }{\alpha \beta \gamma }\right)\\ & ={\left(\frac{a}{-b}\right)}^2-2\cdot \left(\frac{0}{-b}\right)=\frac{a^2}{b^2}-0=\frac{a^2}{b^2}\end{array}$ 

নির্ণেয় মান : $\frac{a^2}{b^2}$ 

দেওয়া আছে, $Q\left(x\right)=b+\mathrm{cx}+{\mathrm{ax}}^2$
এবং $\{Q{\left({\omega }^4\right\}}^3+\{Q{\left(\left({\omega }^2\right)\right\}}^3=0$
বা, ${(b+c{\omega }^4+a{\omega }^8)}^3+{(b+c{\omega }^2+a{\omega }^4)}^3=0$
বা, ${\{a{\left({\omega }^3\right)}^2\cdot {\omega }^2+b+c{\omega }^3\cdot \omega \}}^3+{(a{\omega }^3\cdot \omega +b+c{\omega }^2)}^3$
বা, ${(a{\omega }^2+b+c\omega )}^3+{(a\omega +b+c{\omega }^2)}^3=0[\because {\omega }^3=1]$
ধরি, $p=a{\omega }^2+b+c\omega$ এবং $q=a\omega +b+c{\omega }^2$$\therefore p^3+q^3=0$
বা, $(p+q)(p^2-pq+q^2)=0$
বা, $(p+q)\{p^2+(\omega +{\omega }^2)pq+q^2\cdot {\omega }^3\}=0$

$[\because 1+\omega +{\omega }^2=0$ এবং ${\omega }^3=1]$

বা, $(p+q)(p^2+\omega pq+{\omega }^{2}pq+q^2{\omega }^3)=0$
বা. $(p+q)\left\{p\right(p+q\omega )+q{\omega }^2(p+q\omega \}=0$
বা, $(p+q)(p+q\omega )(p+q{\omega }^2)=0$
বা, $(a{\omega }^2+b+c\omega +a\omega +b+c{\omega }^2)(a{\omega }^2+b+c\omega +a{\omega }^2+b\omega +c{\omega }^3)$

$(a{\omega }^2+b+c\omega +a{\omega }^3+b{\omega }^3+c{\omega }^4)$

বা, $\{a({\omega }^2+\omega )+2b+c(\omega +{\omega }^2)\}\{2a{\omega }^2+b(1+\omega )+c(1+\omega \left)\right\}$

$\{a({\omega }^2+1)+b(1+{\omega }^2)+2c\omega \}=0[\because {\omega }^3=1]$

বা, $(-a+2b-c)(2a{\omega }^2-b{\omega }^2-c{\omega }^2)(-a\omega -b\omega +2c\omega )=0$

$[\because 1+\omega +{\omega }^2=0]$

বা, $(-a+2b-c){\omega }^2(2a-b-c)\omega (2c-a-b)=0$
বা, $(2a-b-c)(2b-a-c)(2c-a-b)=0[\because {\omega }^3=1]$

হয়, $2a-b-c=0$
বা, $2a=b+c$ 

$\therefore a=\frac{b+c}{2}$ 

অথবা, $2b-a-c=0$
বা, $2b=c+a$ 
$\therefore b=\frac{c+a}{2}$ 

অথবা, $2c-a-b=0$
বা, $2c=a+b$
বা, $c=\frac{a+b}{2}$ 

বামপক্ষ $=8abc=8\times \frac{b+c}{2}\times \frac{c+a}{2}\times \frac{a+b}{2}$ 

$=(a+b)(b+c)(c+a)=$ ডানপক্ষ 

$\therefore 8abc=(a+b)(b+c)(c+a)$. (দেখানো হলো)