Question

If $x = 2 + \sqrt{3}$, find the value of ${x}^{3} + \frac{1}{ {x}^{3} }$
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Answer 1

Answer:-

$x = 2 + \sqrt{3}$

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

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

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

$\tt\bold\red\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: =>\frac{1}{x} = 2 - \sqrt{3}$

Therefore,

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

$x + \frac{1}{x} = 2 + 2$

$\tt\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =>x + \frac{1}{x} = 4$

Cubing the both sides as per the question,

${(x + \frac{1}{x}) }^{3} = {4}^{3}$

${x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x}) = 64$

${x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 4 = 64$

${x}^{3} + \frac{1}{ {x}^{3} } + 12 = 64$

${x}^{3} + \frac{1}{ {x}^{3} } = 64 - 12$

$\tt\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =>{x}^{3} + \frac{1}{ {x}^{3} } = 52$

Answer 2

Answer:

Answer:-

x = 2 + \sqrt{3}x=2+3

\frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }x1=2+31×2−32−3

\frac{1}{x} = \frac{2 - \sqrt{3} }{ {2}^{2} - { \sqrt{3} }^{2} }x1=22−322−3

\frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3}x1=4−32−3

\tt\bold\red\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: = > \frac{1}{x} = 2 - \sqrt{3} =>x1=2−3

Therefore,

x + \frac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3}x+x1=2+3+2−3

x + \frac{1}{x} = 2 + 2x+x1=2+2

\tt\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > x + \frac{1}{x} = 4 =>x+x1=4

Cubing the both sides as per the question,

{(x + \frac{1}{x}) }^{3} = {4}^{3}(x+x1)3=43

{x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x}) = 64x3+x31+3(x+x1)=64

{x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 4 = 64x3+x31+3×4=64

{x}^{3} + \frac{1}{ {x}^{3} } + 12 = 64x3+x31+12=64

{x}^{3} + \frac{1}{ {x}^{3} } = 64 - 12x3+x31=64−12

\tt\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > {x}^{3} + \frac{1}{ {x}^{3} } = 52 =>x3+x31=52