Math, asked by roynandini2412, 27 days ago

if x = 2 +root 3 , then find the value of x^3 + 1/x^3

Answers

Answered by sanjey1245
0

Answer:

Answer

It is given that,

x=2−  

3

 

so,

1/x=1/(2−  

3

)

By rationalizing the denominator, we get

=[1(2+  

3

)]/[(2−  

3

)(2+  

3

)]

=[(2+  

3

)]/[(2  

2

)−(  

3

)  

2

]

=[(2+  

3

)]/[4−3]

=2+  

3

 

Now,

x−1/x=2−  

3

−2−  

3

 

=−2  

3

 

Let us cube on both sides, we get

(x−1/x)  

3

=(−2  

3

)  

3

 

x  

3

−1/x  

3

−3(x)(1/x)(x−1/x)=24  

3

 

x  

3

−1/x  

3

−3(−2/  

3

)=−24  

3

 

x  

3

−1/x  

3

+6  

3

=−24  

3

 

x  

3

−1/x  

3

+6  

3

=−24  

3

 

x  

3

−1/x  

3

=−24  

3

−6  

3

 

=−30  

3

 

Hence,

x  

3

−1/x  

3

=−30  

3

 

Step-by-step explanation:

Answered by sandy1816
9

x = 2 +  \sqrt{3}  \\  \frac{1}{x}  =  \frac{1}{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} }{4 - 3}  \\  \frac{1}{x}  = 2 -  \sqrt{3}  \\  \\  \therefore \:  \:  \:  \:  \:  \: x +  \frac{1}{x}  = 4 \\  \\ finding \:  \:  \:  {x}^{3}  +  \frac{1}{ {x}^{3} }  = ( {x +  \frac{1}{x} })^{3}  - 3(x +  \frac{1}{x} ) \\  =  {4}^{3}  - 3.4 \\  = 64 - 12 \\  = 52

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