Question

If x = (2+under root 3),show that (x cube +1/x cube )=52

Answer 1

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies x = 2 + \sqrt{3} \\ \\ \red{\underline \bold{To \: Show : }} \\ \tt: \implies {x}^{3} + { (\frac{1}{x} )}^{3} = 52$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {x}^{3} = {(2 + \sqrt{3} )}^{2}\\\\ \tt\circ\:(a+b)^{3}=a^{3}+b^{3}+3a^{2}b+3ab^{2} \\ \\ \tt: \implies {x}^{3} = {2}^{3} + \sqrt{3}^{3} + 3 \times {2}^{2} \times \sqrt{3} + 3 \times 2 \times { \sqrt{3} }^{2} \\ \\ \tt: \implies {x}^{3} =8 + 3 \sqrt{3} + 12 \sqrt{3} + 18 \\ \\ \tt: \implies {x}^{3} =26 + 15 \sqrt{3} - - - - - (1) \\ \\ \bold{As \:we \: know \: that} \\ \tt: \implies (\frac{1}{x} )^{3} = ( \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} })^{3} \\ \\ \tt: \implies (\frac{1}{x} )^{3} = (\frac{2 - \sqrt{3} }{4 - 3} )^{3} \\ \\ \tt: \implies (\frac{1}{x} )^{3} = {(2 - \sqrt{3} )}^{3} \\\\ \tt\circ\:(a-b)^{3}=a^{3}-b^{3}-3a^{2}b+3ab^{3} \\ \\ \tt: \implies (\frac{1}{x} )^{3} = {2}^{3} - (\sqrt{3} )^{2} - 3 \times {2}^{2} \times \sqrt{3} + 3 \times 2 \times (\sqrt{3})^{2} \\ \\ \tt: \implies (\frac{1}{x} )^{3} =8 - 3 \sqrt{3} - 12 \sqrt{3} + 18 \\ \\ \tt: \implies (\frac{1}{x} )^{3} =26 - 15 \sqrt{3} - - - - - (2) \\ \\ \bold{For \: finding \: value : } \\ \tt: \implies {x}^{3} + (\frac{1}{x} )^{3} =26 + 15\sqrt{3} + 26 - 15 \sqrt{3} \\ \\ \green{\tt: \implies {x}^{3} + (\frac{1}{x} )^{3} =52} \\ \\ \green{\huge{\tt \:\:\:\:Proved }}$

Answer 2

$x = 2 + \sqrt{3}$

$\frac{1}{x} = \frac{2 - \sqrt{3} }{(2 + \sqrt{3) \times (2 - \sqrt{3) } } }$

$\frac{1}{x } = \frac{2 - \sqrt{3} }{4 - 3}$

$\frac{1}{x} = 2 - \sqrt{3}$

${x}^{3} + \frac{1}{ {x}^{3} } = {(x + \frac{1}{x}) }^{3} - 3 \times x \times \frac{1}{x} (x + \frac{1}{x} )$

${(2 + \sqrt{3 } + 2 - \sqrt{3}) }^{3} - 3(2 + \sqrt{3} + 2 - \sqrt{3}$

${4}^{3} - 3(4)$

$64 - 12 = 52 \: proved$