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# stuck on this equation

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

Jul 12, 2020

#1
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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