Question

$find \: the \: value \: of \\ (x3 - \frac{1}{ {x}^{3} } )if \frac{x}{2} = \frac{1 }{2x} + 1$
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Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:\dfrac{x}{2} = \dfrac{1}{2x} + 1$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} }$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:\dfrac{x}{2} = \dfrac{1}{2x} + 1$

On multiply by 2, we get

$\rm :\longmapsto\:x = \dfrac{1}{x} + 2$

can be rewritten as

$\rm :\longmapsto\:x - \dfrac{1}{x} = 2$

Now,

Consider,

$\rm :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} }$

$\rm \:  =  \:  \: {\bigg(x - \dfrac{1}{x} \bigg) }^{3} - 3 \times x \times \dfrac{1}{x}\bigg(x - \dfrac{1}{x} \bigg)$

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

$\rm \:  =  \:  \: {(2)}^{3} - 3 \times 2$

$\rm \:  =  \:  \: 8 - 6$

$\rm \:  =  \:  \: 2$

Hence,

$\boxed{ \bf{ \: \: \bf :\longmapsto\: {x}^{3} - \dfrac{1}{ {x}^{3} } = 2 \: \: \: \: }}$

Additional Information :-

More Identities to know : -

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b² = (a + b)(a - b)

(a + b)² = (a - b)² + 4ab

(a - b)² = (a + b)² - 4ab

(a + b)² + (a - b)² = 2(a² + b²)

(a + b)³ = a³ + b³ + 3ab(a + b)

(a - b)³ = a³ - b³ - 3ab(a - b)