Question

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Answer 1

Given Question :-

$\sf \: If \: {a}^{2} + \dfrac{1}{ {a}^{2} } = 27, \: find \: the \: value \: of \: a - \dfrac{1}{a}$

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

Given that,

$\rm :\longmapsto\: {a}^{2} + \dfrac{1}{ {a}^{2} } = 27$

can be rewritten as

$\rm :\longmapsto\: {(a)}^{2} + {\bigg[\dfrac{1}{a} \bigg]}^{2} = 27$

On Subtracting 2 from each term, we get

$\rm :\longmapsto\: {(a)}^{2} + {\bigg[\dfrac{1}{a} \bigg]}^{2} - 2 = 27 - 2$

$\rm :\longmapsto\: {(a)}^{2} + {\bigg[\dfrac{1}{a} \bigg]}^{2} - 2 \times a \times \dfrac{1}{a} = 25$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: {x}^{2} - 2xy + {y}^{2} = {(x - y}^{2} \: }}}$

So, using this, we get

$\rm :\longmapsto\: {\bigg[a - \dfrac{1}{a} \bigg]}^{2} = 25$

$\rm :\longmapsto\: {\bigg[a - \dfrac{1}{a} \bigg]}^{2} = 5 \times 5$

$\rm :\longmapsto\: {\bigg[a - \dfrac{1}{a} \bigg]}^{2} = {5}^{2}$

$\rm \implies\:\boxed{ \tt{ \: a - \frac{1}{a} = \: \pm \: 5 \: }}$

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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)

Answer 2

Answer:

Option (a) is correct...