Question

if X square + 1 by x square equals to 23 then find the value of x cube + 1 by x cube​

Answer 1

Step-by-step explanation:

${x}^{2} + \frac{1}{ {x}^{2}} = 23 = > {x}^{2} + \frac{1}{ {x}^{2} } + 2.x. \frac{1}{x} =$

$25 = > {(x + \frac{1}{x})}^{2} = 25 = > (x + \frac{1}{x}) =$

$5$

${x}^3 + {y}^{3} = (x + y)( {x}^{2} + {y}^{2} - xy)$

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

$= > 5(23 - 1) = > 110$

Hope it helps you.

Answer 2

The value of $x^3+\dfrac{1}{x^3}$ is 110.

 Explanation:

Given :   $x^2+\dfrac{1}{x^2}=23$

Consider :  

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

    [∵ (a+b)²= a²+b²+2ab]

$=x^2+\dfrac{1}{x^2}+2$



$=23+2=25$                  (Substitute value of   $x^2+\dfrac{1}{x^2}=23$)

$\Rightarrow\ (x+\dfrac{1}{x})^2=25\\\\\Rightarrow\ (x+\dfrac{1}{x})=\sqrt{25}=5$

Consider : $x^3+\dfrac{1}{x^3}=(x+\dfrac{1}{x})(x^2+\dfrac{1}{x^2}-x(\dfrac{1}{x}))$   [∵ a³+b³=(a+b)(a²+b²-ab)]

$=(5)(23-1)=5(22)=110$  (Substitute value of   $x^2+\dfrac{1}{x^2}=23$ and $x+\dfrac{1}{x}=5$)

Hence, the value of $x^3+\dfrac{1}{x^3}$ is 110.

# Learn more :

If a + b is equals to 3 and a square + b square is equal to 40 find the value of a cube plus b cube​

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