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

\((m^2+\frac{1}{m^2} )^2= 121\)

\((m^2)^2 + 2m^2\cdot\frac{1}{m^2} + \left(\frac{1}{m^2}\right)^{\!2} = 121\)

\(m^4+2+\frac{1}{m^4}=121\)

\(m^4+\frac{1}{m^4}=\boxed{119}\)

.