It is given \(m+\dfrac{1}{m}=4.\)Find the value of \(m^4+\dfrac{1}{m^4}\)

We have :

m4+1m4=m4+2+1m4−2=(m2+1m2)2−2

=(m2+2+1m2−2)2−2=[(m+1m)2−2]2−2

=(42−2)2−2=142−2=194

\(m^4+\dfrac{1}{m^4}=m^4+2+\dfrac{1}{m^4}-2=\left(m^2+\dfrac{1}{m^2}\right)^2-2\)

\(=\left(m^2+2+\dfrac{1}{m^2}-2\right)^2-2=\left[\left(m+\dfrac{1}{m}\right)^2-2\right]^2-2\)

\(=\left(4^2-2\right)^2-2=14^2-2=194\)