+738 hi guest!
so we can start by raising \(m-\frac{1}{m}=3\) to the 2nd power.
\((m-\frac{1}{m})^2=(3)^2\)
\(m^2 + 2m\cdot \frac{1}{m} + \frac{1}{m^2} = 9\)
\(m^2 -2+ \frac{1}{m^2} = 9\)
\(m^2+\frac{1}{m^2} = 11\)
now let's square this again
\((m^2+\frac{1}{m^2} )^2= 121\)
\((m^2)^2 + 2m^2\cdot\frac{1}{m^2} + \left(\frac{1}{m^2}\right)^{\!2} = 121\)
\(m^4+2+\frac{1}{m^4}=121\)
\(m^4+\frac{1}{m^4}=\boxed{119}\)
