Question

if
$m + \frac{1}{m} = 4$
then what is the value of
$m3 + \frac{1}{ {m}^{3} }$
​

Answer 1

$\large \bf{Given...}$

Given that $m+\dfrac{1}{m}=4$.

$\large \bf{To~ find...}$

We have to find $m^3+\dfrac{1}{m}$.

$\large\bf{Solution...}$

• We have the equation :

$\: \: \: \: : \implies m + \dfrac{1}{m} = 4$

• Now, we'll cube it on both sides :

$\: \: \: \: : \implies {(m + \dfrac{1}{m} })^{3} = {(4)}^{3}$

• Using the identity : $a^3+b^3=(a^3+b^3)+3ab(a+b)$ :

$\: \: \: \: : \implies {m}^{3} + { \dfrac{1}{m} }^{3} + 3(m \times \dfrac{1}{m} )(m + \dfrac{1}{m} ) = 64$

$\: \: \: \: : \implies {m}^{3} + \dfrac{1}{ {m}^{3} } + 3( \cancel m + \dfrac{1}{ \cancel m} )(4) = 64 \: \: \: \: \: \: \: \bf{( \because \: m + \dfrac{1}{m} = 4)}$

$\: \: \: \: : \implies {m}^{3} + \dfrac{1}{ {m}^{3} } + 3(1)(4) = 64$

$\: \: \: \: : \implies {m}^{3} + \dfrac{1}{ {m}^{3} } + 12 = 64$

$\: \: \: \: : \implies {m}^{3} + \dfrac{1}{ {m}^{3} } = 64 - 12$

$\: \: \: \: : \implies {m}^{3} + \dfrac{1}{ {m}^{3} } = 52$

_______________________

$\bf {Therefore,~ m^3+\dfrac{1}{m^3}=52,~when~ m+\dfrac{1}{m}=4}.$