Question

If $x+\frac{1}{x} =3$, then find the value of $x^{3}+\frac{1}{x^3}$

Answer 1

$\large\underline{\sf{We \ have}}$

$\sf{x - \frac{1}{x} = 3}$

$\large\underline{\sf{To \ find}}$

$\sf{the \ value \ of {x}^{3} - \frac{1}{{x}^{3}}}$

$\large\underline{\sf{On \ cubing \ both \ side}}$

$\sf{{(x - \frac{1}{x})}^{3} = {(3)}^{3}}$

$\large\underline{\sf{Using \ identity}}$

$\fbox{\sf{(a - b)³ = a³ - b³ - 3ab (a - b)}}$

$\sf{x³ - \frac{1}{x³} - 3 (x - \frac{1}{x}) = 27}$

$\sf{x³ - \frac{1}{x³} - 3(3) = 27}$

$\sf{x³ - \frac{1}{x³} - 9 = 27}$

$\sf{x³ - \frac{1}{x³} = 27 + 9}$

$\sf{x³ - \frac{1}{x³} = 36}$

The Value of

$\sf\green{x³ - \frac{1}{x³} = 36}$

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Answer 2

$\large\underline{\rm{We \ Have}}$

$\rm{x - \dfrac{1}{x} = 3}$

$\large\underline{\rm{To \ Find}}$

$\rm{x³ - \dfrac{1}{x³}}$

$\large\underline{\rm{On \ cubing \ both \ side}}$

$\rm{(x - \dfrac{1}{x})³ = (3)³}$

$\large\underline{\rm{Using \ identity}}$

$\fbox{\rm{(a - b)³ = a³ - b³ - 3ab (a - b)}}$

$\rm{x³ - \dfrac{1}{x³} - 3 (x - \dfrac{1}{x}) = 27}$

$\rm{x³ - \dfrac{1}{x³} - 3(3) = 27}$

$\rm{x³ - \dfrac{1}{x³} - 9 = 27}$

$\rm{x³ - \dfrac{1}{x³} = 27 + 9}$

$\rm{x³ - \dfrac{1}{x³} = 36}$

The Value of

$\red{\boxed{\rm{x³ - \frac{1}{x³} = 36}}}$

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