Question

If x+1/x=2 then find the value of x
${x}^{3 } + 1 \frac{1}{?x {?}^{3} }$

Answer 1

$<b >Hey here is your answer$

$= x + \frac{1}{x} = 2 \\ \\ = squaring \: both \: side \: \\ \\ = (x + \frac{1}{x} {)}^{2} = {(2)}^{2}$

using identity

${x}^{2} + { \frac{1}{x} }^{2} + 2 \times x \times \frac{1}{x} = 4 \\ \\ = {x}^{2} + { \frac{1}{x} }^{2} + 2 = 4 \\ \\ = {x}^{2} + { \frac{1}{x} }^{2} =4 - 2 = 2 \\ \\ \\ = {x}^{3} + { \frac{1}{x} }^{3}$

using identity

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

by substituting the values

$2(2 - 1) = 2 \times 1 \\ \\ = 2$

hope this helps you

Answer 2

Hi , It is given that ,  

x + 1/x = 2 ----( 1 )  *********************************

 We know the algebraic identity:

 a³ + b³ + 3ab(a + b) = (a + b)³   ************************************
 Here ,.  

a = x , b = 1/x  

x³+1/x³+3×x×1/x (x+1/x) = (x+1/x)³

 => x³ + 1/x³ + 3(x + 1/x ) = ( x + 1/x )³  

=>x³ + 1/x³ + 3 × 2 = 2³ [ from ( 1 ) ]

 x³ + 1/x³ + 6 = 8

 x³ + 1/x³ = 8 - 6    

= 2  

I hope this helps you.

 : )