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If \(m - \frac{1}{m} = 3\) then find \(m^4 + \frac{1}{m^4}\)

 Jul 12, 2020
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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

 

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

 

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

 

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

 

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

 Jul 13, 2020

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