Math, asked by 0202sushiladubey, 10 months ago

if (x+1/x)2=3 , shows that (x3+1/x3) 0​

Answers

Answered by Anonymous
52

Answer:

Proved.

Step-by-step explanation:

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Answered by rinayjainsl
8

Answer:

Our required relation is obtained and it is

{x}^{3} +  \frac{1}{ {x}^{3} } = 0

Step-by-step explanation:

The given algebraic relation of second degree is

(x +  \frac{1}{x} ) {}^{2}  = 3

Squaring and expanding the terms on left side,we get

 {x}^{2}  +  { \frac{1}{x {}^{2} } } + 2x( \frac{1}{x} ) = 3 \\  =  >  {x}^{2}  +  \frac{1}{ {x}^{2} }  + 2 = 3 \\  =  > {x}^{2}  +  \frac{1}{ {x}^{2} } = 1 -  -  > (1)

Now for obtaining our desired output relation,we cube the relation on both sides.The cubing of equation is done based on the identity as shown below.

(a + b) {}^{3}  =  {a}^{3}  +  {b}^{3} + 3ab(a + b)

Hence,cubing the equation 1 on both sides,we get

 \\  ({x}^{2} ) {}^{3}  + ( \frac{1}{ {x}^{2} } ) {}^{3}  + 3 {x}^{2} ( \frac{1}{ {x}^{2} } )( {x}^{2}  +  \frac{1}{{x}^{2}  }) =  {1}^{3}  \\  =  >  {x}^{6}  +  \frac{1}{ {x}^{6} }  + 3( {x}^{2}  +  \frac{1}{ {x}^{2} } ) = 1

Substituting the value of equation (1) in the above relation,we get

 {x}^{6}  +  \frac{1}{ {x}^{6} }  + 3 = 1 \\  =  >  {x}^{6}  +  \frac{1}{ {x}^{6} }   + 2 = 0

Now we shall rearrange a equation in such a way that we can write the entire term as a square of a single term.This can be done in thr following way as shown-

( {x}^{3} ) {}^{2}  + ( \frac{1}{ {x}^{3} } ) {}^{2}  + 2( {x}^{3} )( \frac{1}{ {x}^{3} } ) = 0 \\

The above relation is of the form as shown below-

 {a}^{2}  +  {b}^{2}  + 2ab =  {(a + b)}^{2}

Hence we can rewrite the above term as

( {x}^{3} +  \frac{1}{ {x}^{3} }  ) {}^{2}  = 0 \\  =  > {x}^{3} +  \frac{1}{ {x}^{3} } = 0

Hence,the required relation is obtained.

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