Question

If equation $\left(l^2+m+n+1\right)x^2+\left(m^2+n+l+1\right)x+\left(n^2+l+m+1\right)=0$ and $(q-r)x^2+(r-p)x+(p-q)=0$ have a common root and $2^{\text{nd }}$ equation has equal roots then find the value of $l^{2016}+m^{2016}+n^{2016}-\frac{6}{l^{2015}+m^{2015}+n^{2015}}$ where $l,m,n,p,q,r\in R$.

Answer 1

Answer: 5

Clearly one root of equation $(q-r)x^2+(r-p)x+(p-q)=0$ is 1 and both roots are equal.
$\therefore$ both roots are equal to 1 .
$\therefore 1$ will also be a root of $\left(1^2+m+n+1\right)x^2+\left(m^2+n+l+1\right)x+\left(n^2+l+m+1\right)=0$
$\therefore l^2+m+n+1+m^2+n+l+1+n^2+l+m+1=0$
$\Rightarrow (1+1{)}^2+(m+1{)}^2+(n+1{)}^2=0$
$\Rightarrow l=m=n=-1$
$\therefore$ Given expression $=1+1+1-\frac{6}{(-1)+(-1)+(-1)}=3+2=5$