Question

If x+1/x = underroot 3. Find the value of x3+1/x3

Answer 1

$\mathfrak{\large{\underline{\underline{Correct \: Question:-}}}}$

If x + 1/x = √3. Find the value of x³ + 1/x³

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{ \tt {x}^{3} + \dfrac{1}{ {x}^{3}} = 0}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\tt x + \dfrac{1}{x} = \sqrt{3}$

To find :- $\tt x^3 + \dfrac{1}{x^3}$

Solution :-

$\tt x + \dfrac{1}{x} = \sqrt{3}$

Cubing on both sides

$\tt (x + \dfrac{1}{x})^{3} = (\sqrt{3})^{3}$

We know that (x + y)³ = x³ + y³ + 3xy(x + y)

Here x = x, y = 1/x

By substituting the values in the identity we have

$\tt {x}^{3} + \dfrac{1}{ {x}^{3} } + 3(x)( \dfrac{1}{x})(x + \dfrac{1}{x}) = 3 \sqrt{3}$

$\tt {x}^{3} + \dfrac{1}{ {x}^{3}} + 3(x + \dfrac{1}{x}) = 3 \sqrt{3}$

$\tt {x}^{3} + \dfrac{1}{ {x}^{3}} + 3( \sqrt{3}) = 3 \sqrt{3}$

[Since given that $\bf x + \dfrac{1}{x} = \sqrt{3}$ ]

$\tt {x}^{3} + \dfrac{1}{ {x}^{3}} + 3 \sqrt{3} = 3 \sqrt{3}$

$\tt {x}^{3} + \dfrac{1}{ {x}^{3}}= 3 \sqrt{3} - 3 \sqrt{3}$

$\tt {x}^{3} + \dfrac{1}{ {x}^{3}} = 0$

$\Huge{\boxed{ \sf {x}^{3} + \dfrac{1}{ {x}^{3}} = 0}}$

Answer 2

x + $\dfrac{1}{x}$ = √3

________ [GIVEN]

• We have to find the value of x³ + $\dfrac{1}{ {x}^{3} }$

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=> x + $\dfrac{1}{x}$ = √3

• Take cube on both sides..

=> $( {x \: + \: \dfrac{1}{x}) }^{3}$ = (√3)³

=> x³ + $\dfrac{1}{ {x}^{3} }$ + 3x × $\dfrax{1}{x}$ $(x \: + \: \dfrac{1}{x})$ = 3√3

=> x³ + $\dfrac{1}{ {x}^{3} }$ + 3(√3) = 3√3

=> x³ + $\dfrac{1}{ {x}^{3} }$ = 3√3 - 3√3

______________________________

x³ + $\dfrac{1}{ {x}^{3} }$ = 0

___________ [$\bold{ANSWER}]$

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✡ Used identity :

(a + b)³ = a³ + b³ + 3ab(a + b)

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