Question

If $m-\frac{1}{m}=5$, then find $m^4+\frac{1}{m^4}$

Answer 1

Applying the formula, ${(a-b)}^2=a^2+b^2-2ab$
${(m-\frac{1}{m})}^2=m^2+{\left(\frac{1}{m}\right)}^2-2m\times \frac{1}{m}=m^2+\frac{1}{m^2}-2$
$=>m^2+\frac{1}{m^2}={(m-\frac{1}{m})}^2+2$
Substituting $m-\frac{1}{m}=5$,
$=>m^2+\frac{1}{m^2}=5^2+2=25+2=27$

Squaring both sides,

${(m^2+\frac{1}{m^2})}^2=27\times 27=729$

$m^4+\frac{1}{m^4}+2\times m^2\times \left(\frac{1}{m^2}\right)=729$

$m^4+\frac{1}{m^4}+2=729$
$m^4+\frac{1}{m^4}=727$