Question

if X square + 1 upon x square is equal to 7 find the value of x cube + 1 / x cube

Answer 1

Given

x^2 + 1/x^2 = 7

To Find

x^3 + 1/x^3 = ?

Solution

( x + 1/x )^2 = ( x )^2 + ( 1/x )^2 + 2 ( x )( 1/x )

( x + 1/x )^2 = x^2 + 1/x^2 + 2

( x + 1/x )^2 = 7 + 2

( x + 1/x )^2 = 9

x + 1/x = 3

( x + 1/x )^3 = ( x )^3 + ( 1/x )^3 + 3 ( x )( 1/x ) ( X + 1/x )

( 3 )^3 = x^3 + 1/x^3 + 3 ( 3 )

27 = x^3 + 1/x^3 + 9

x^3 + 1/x^3 = 18

Answer 2

Hey friend, Harish here.
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Here is your answer
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Given that,

$x^{2} + \frac{1}{ x^{2} } = 7$
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To Find,

$x^{3} + \frac{1}{ x^{3} } = ?$
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Solution,

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

Add, 2 on both the sides.
 
⇒$x^{2} + \frac{1}{ x^{2} } + 2 =7+ 2$

⇒$( x+ \frac{1}{ x} )^{2} = 9$   (Take square root on both the sides)

⇒$(x+ \frac{1}{x} ) = 3$  - (i)
 
Now cube on both sides in equation (i).

⇒$(x+ \frac{1}{x} )^{3} = 3^{3}$

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

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

Now, Substitute the value of $x + \frac{1}{x}$ in th above equation.

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

⇒$x^{3}+ \frac{1}{x^{3}} = 27 - 9$

⇒$x^{3}+ \frac{1}{x^{3}} = 18$
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Hope my answer is helpful to u.