Question

if x^2+1/x^2=7, then find the valuex^3+1/x^3=?​

Answer 1

Heya!!

x² + 1/x² = 7 Find x³+1/x³

( x + 1/x )² - 2 = 7

Becoz ( a² + b² ) = ( a + b )² - 2ab

( x + 1/x )² = 9

( x + 1/x ) = ±3

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

Becoz ( a³ + b³ ) = ( a + b )³ -3ab ( a + b )

x³ + 1/x³ = ( ±3 )³ - 3 ( ± 3 )

Taking -ive values

x³ + 1/x³ = -27 + 9

Taking +ive values

x³ + 1/x³ = 27 - 9

x³ + 1/x³ = -18

OR

x³ + 1/x³ = 18

Answer 2

Given, $\: \sf{ {x}^{2} + \frac{1}{ {x}^{2} } = 7 }$

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

$\sf {( x + \frac{1}{x} )}^{2} = 7 + 2$

$\sf {( x + \frac{1}{x} )}^{2} = 9$

$\sf x + \frac{1}{x} = \sqrt{9}$

$\sf x + \frac{1}{x} = 3$

Now,

Cubing on both sides

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

$\sf { {x}^{3} + \frac{1}{ {x}^{3} }} + 3(x + \frac{1}{x}) = 27$

$\sf { {x}^{3} + \frac{1}{ {x}^{3} }} + 3(3) = 27$

$\sf { {x}^{3} + \frac{1}{ {x}^{3} }} = 27 - 9$

$\fbox{\sf { {x}^{3} + \frac{1}{ {x}^{3} }} = 18}$