সৃজনশীলঃ ‘F(l, m,n)=l3+m3+n3−3lmn একটি…’ ২০২৫ – বীজগাণিতিক রাশি, উচ্চতর গণিত (এসএসসি)

এখানে, $F(l,m,n)=0$ এবং $l+m+n\ne 0$ 

$\begin{array}{cc}& F(l,m,n)=l^3+m^3+n^3-3l\mathrm{mn}\\ & =(l+m{)}^3-3lm(l+m)+n^3-3l\mathrm{mn}\\ & =(l+m{)}^3+n^3-3lm(l+m)-3l\mathrm{mn}\\ & =(l+m+n)\left\{\right(l+m{)}^2-(l+m)n+n^2\}-3lm(l+m+n)\\ & =(l+m+n)(l^2+2lm+m^2-ln-\mathrm{mn}+n^2-3lm)\\ & =(l+m+n)(l^2+m^2+n^2-lm-\mathrm{mn}-nl)\end{array}$ 

যেহেতু$F(l,m,n)=0$
সেহেতু $(l+m+n)(l^2+m^2+n^2-lm-\mathrm{mn}-nl)=0$ 
বা, $l^2+m^2+n^2-lm-\mathrm{mn}-nl=0[\because l+m+n\ne 0]$ 

$\therefore l^2+m^2+n^2=lm+\mathrm{mn}+\mathrm{nl}$. (দেখানো হলো) 

দেওয়া আছে, $F(l,m,n)=l^3+m^3+n^3-3l\mathrm{mn}$. 
এখানে, $l=y+z-x,m=z+x-y$ এবং, $n=x+y-z$ 

$\therefore \begin{array}{cc}& F(l,m,n)=\frac{1}{2}(l+m+n)\left\{\right(l-m{)}^2+(m-n{)}^2+(n-l{)}^2\}\\ =& \frac{1}{2}(y+z-x+z+x-y+x+y-z)\\ & \left\{\right(y+z-x-z-x+y{)}^2+(z+x-y-x-y+z{)}^2+(x+y-z-y-z+x{)}^2\}\\ =& \frac{1}{2}(x+y+z)\left\{\right(2y-2x{)}^2+(2z-2y{)}^2+(2x-2z{)}^2\}\\ =& \frac{1}{2}(x+y+z)\left\{4\right(x-y{)}^2+4(y-z{)}^2+4(z-x{)}^2\}\\ =& 4\frac{1}{2}(x+y+z)\left\{\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2\}\\ =& 4(x^3+y^3+z^3-3\mathrm{xyz})=4F(x,y,z)\end{array}$ 

অর্থাৎ, $4F(x,y,z)=F(l,m,n)$. (দেখানো হলো)