stuck on this equation

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

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

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

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

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