সৃজনশীলঃ ‘দৃশ্যকল্প-১ : p(x)=(a+b+c)x2+(b+2c)x+c;…’ নটরডেম কলেজ ২০২৪ – বহুপদী ও বহুপদী সমীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ২য় পত্র (এইচএসসি)

দেওয়া আছে, $p\left(x\right)=(a+b+c)x^2+(b+2c)x+c$এবং $p\left(x\right)=0$

বা, $(a+b+c)x^2+(b+2c)x+c=0$ এর মূলদ্বয় $\alpha ,\beta$

$\therefore \alpha +\beta =\frac{-(b+2c)}{a+b+c}$এবং $\alpha \beta =\frac{c}{a+b+c}$

এখন, মূলদ্বয়ের যোগফল$=\frac{\alpha }{\alpha +1}+\frac{\beta }{\beta +1}=\frac{\alpha \beta +\alpha +\alpha \beta +\beta }{(\alpha +1)(\beta +1)}$

$=\frac{(\alpha +\beta )+2\alpha \beta }{\alpha \beta +\alpha +\beta +1}=\frac{\frac{-(b+2c)}{a+b+c}+\frac{2c}{a+b+c}}{\frac{c}{a+b+c}+\frac{-(b+2c)}{a+b+c}+1}$

$ =frac{-b-2 c+2 c}{c-b-2 c+a+b+c}=frac{-b}{a}

এবং মূলদ্বয়ের গুণফল $=\frac{\alpha }{\alpha +1}\cdot \frac{\beta }{\beta +1}=\frac{\alpha \beta }{\alpha \beta +\alpha +\beta +1}$

$=\frac{\frac{c}{a+b+c}}{\frac{c}{a+b+c}+\frac{-(b+2c)}{a+b+c}+1}$

$ =frac{c}{c-b-2 c+a+b+c}=frac{c}{a}

$\therefore \frac{\alpha }{\alpha +1}\otimes \frac{\beta }{\beta +1}$মূলবিশিষ্ট সমীকরণ,

$x^2-($ মূলদ্বয়ের যোগফল) x+ মূলদ্বয়ের গুণফল $=0$

বা, $x^2-\frac{-b}{a}x+\frac{c}{a}=0$

$\therefore a{x}^2+bx+c=0$

নির্ণেয় সমীকরণ$a^2+bx+c=0$.

বামপক্ষ$=x^3+y^3+z^3-3xyz$

$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

$=(x+y+z)\{x^2+y^2+z^2+(\omega +{\omega }^2)xy$

$+(\omega +{\omega }^2)yz+(\omega +{\omega }^2)zx\}$

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

$=(x+y+z)\{x^2+y^2+z^2+\omega xy+{\omega }^{2}xy+\omega yz+{\omega }^{2}yz+\omega zx+{\omega }^{2}zx\}$

$=(x+y+z)\{x^2+{\omega }^{2}xy+\omega zx+\omega xy+{\omega }^3{y}^2$

$+{\omega }^{4}yz+{\omega }^{2}zx+{\omega }^{2}yz+{\omega }^3{z}^2\}[\because {\omega }^3=1]$

$=(x+y+z)\{x(x+{\omega }^{2}y+\omega z)+\omega y$

$(x+{\omega }^{2}y+\omega z)+{\omega }^{2}z(x+{\omega }^{2}y+\omega z)\}$

$=(x+y+z)(x+{\omega }^{2}y+\omega z)(x+\omega y+{\omega }^{2}z)$

$=(x+y+z)(x+\omega y+{\omega }^{2}z)(x+{\omega }^{2}y+\omega z)$

= ডানপক্ষ

$\therefore x^3+y^3+z^3-3xyz=(x+y+z)(x+\omega y+{\omega }^{2}z)(x+{\omega }^{2}y+\omega z)$. (প্রমাণিত)