Question

x²+1/x² = 38 , then find the value of
x³-1/x³​

Answer 1

Answer:

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Answer 2

${x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: = \: 38$

_____________ [GIVEN]

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

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=> ${x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \: = \: 38$

Subtract 2 on both sides..

=> ${x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \:-\:2 = \: 38\:-\:2$

We can also write it like ...

=> ${x}^{2} \: + \: \dfrac{1}{ {x}^{2} } \:-\:2 (x\:+\:\dfrac{1}{x})\: = \: 38\:-\:2$

(a - b)² = a² + b² - 2ab

=> $\bigg(x \: - \: \dfrac{1}{x} \bigg)^{2} \: = \: 36$

=> $x\:-\:\dfrac{1}{x}\:=\:\sqrt{36}$

=> $x\:-\:\dfrac{1}{x}\:=\:6$ _______ (eq 1)

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• $x\:-\:\dfrac{1}{x}\:=\:6$

Cube on both sides ..

=> $\bigg(x \: - \: \dfrac{1}{x} \bigg)^{3} \: = \: ( {6)}^{3}$

(a - b)³ = a³ - b³ - 3ab (a + b)

=> ${x}^{3} \: - \: \dfrac{1}{ {x}^{3} } \: - \: 3x \: \times \: \dfrac{1}{x} (x \: - \: \dfrac{1}{x} ) \: = \: 216$

=> ${x}^{3} \: - \: \dfrac{1}{ {x}^{3} } \: - \: 3 (6) \: = \: 216$

=> ${x}^{3} \: - \: \dfrac{1}{ {x}^{3} } \: - \: 18 \: = \: 216$

=> ${x}^{3} \: - \: \dfrac{1}{ {x}^{3} } \: = \: 216\:+\:18$

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${x}^{3} \: - \: \dfrac{1}{ {x}^{3} } \: = \: 234$

___________ [ANSWER]

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