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

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 $=1+1+1-\frac{6}{(-1)+(-1)+(-1)}=3+2= 5$