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Ch- Quadratic Surds​

Answer 1

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

$\rm :\longmapsto\:x = \sqrt{3} + \sqrt{2}$

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

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

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\red{\boxed{ \bf{ \: (x + y)(x - y) = {x}^{2} - {y}^{2}}}}$

$\red{\boxed{ \bf{ \: {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y)}}}$

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

Given that

$\rm :\longmapsto\:x = \sqrt{3} + \sqrt{2}$

Consider,

$\rm :\longmapsto\:\dfrac{1}{x}$

$\rm \: = \: \:\dfrac{1}{ \sqrt{3} + \sqrt{2} }$

On rationalizing the denominator, we get

$\rm \: = \: \:\dfrac{1}{ \sqrt{3} + \sqrt{2} } \times \dfrac{ \sqrt{3} - \sqrt{2} }{ \sqrt{3} - \sqrt{2} }$

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

$\rm \: = \: \:\dfrac{ \sqrt{3} - \sqrt{2} }{3 - 2}$

$\rm \: = \: \:\dfrac{ \sqrt{3} - \sqrt{2} }{1}$

$\rm \: = \: \: \sqrt{3} - \sqrt{2}$

$\bf\implies \:\dfrac{1}{x} \: = \: \sqrt{3} - \sqrt{2}$

So,

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

$\rm \: = \: \: \sqrt{3} + \sqrt{2} + \sqrt{3} - \sqrt{2}$

$\rm \: = \: \:2 \sqrt{3}$

$\bf\implies \:\:x + \dfrac{1}{x} = 2 \sqrt{3}$

Now,

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

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

$\rm \: = \: \: {(2 \sqrt{3}) }^{3} - 3 \times 1 \times 2 \sqrt{3}$

$\rm \: = \: \:24 \sqrt{3} - 6 \sqrt{3}$

$\rm \: = \: \:18 \sqrt{3}$

$\bf\implies \: {x}^{3} + \dfrac{1}{ {x}^{3} } = 18 \sqrt{3}$

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)