Math, asked by jaishankartripathi66, 1 month ago

If x = 2-√3 Find the value of (x+1/x)3 ans is -24√3​

Answers

Answered by 24343nagma
0

Step-by-step explanation:

Given :

x = 2 - \sqrt{3}x=2−

3

To find :

x {}^{3} + \frac{1}{x{}^{3}}x

3

+

x

3

1

Solution :

\begin{gathered}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} \\ \\ \frac{1}{x} = 2 + \sqrt{3} \end{gathered}

x=2−

3

x

1

=

2−

3

1

×

2+

3

2+

3

x

1

=

(2)

2

−(

3

)

2

2+

3

x

1

=

4−3

2+

3

x

1

=2+

3

Now,

\begin{gathered}x + \frac{1}{x} = 2 - \cancel {\sqrt{3}} + 2 + \cancel {\sqrt{3}} \\ \\ x + \frac{1}{x} = 2 + 2 \\ \\ x + \frac{1}{x} = 4\end{gathered}

x+

x

1

=2−

3

+2+

3

x+

x

1

=2+2

x+

x

1

=4

And,

On cubing both sides.

\begin{gathered}(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 \\ \\ \boxed{ \bold{ x {}^{3} + \frac{1}{x{}^{3} } = 52}}\end{gathered}

(x+

x

1

)

3

=(4)

3

x

3

+

x

3

1

+3(x+

x

1

)=64

x

3

+

x

3

1

+3×4=64

x

3

+

x

3

1

+12=64

x

3

+

x

3

1

=64−12

x

3

+

x

3

1

=52

_______________________

Thanks for the question!

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