Question

if x square + 1/ x square = 23 , then find the value of x + 1/ x cube
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Answer 1

GiveN :-

• $\sf {x}^{2} + \dfrac{1}{ {x}^{2} } = 23$

To FinD :-

• $\sf {x}^{3} +\dfrac{1}{ {x}^{3}}$

SolutioN :-

$\longrightarrow \sf {x}^{2} + \dfrac{1}{ {x}^{2} } = 23$

Add 2 both side

$\longrightarrow \sf {x}^{2}+\frac{1}{{x}^{2}} + 2 = 23 + 2 \\ \\ \longrightarrow \sf {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} = 25 \\ \\ \longrightarrow \sf{ \bigg(x + \frac{1}{x} \bigg) }^{2} = 25 \\ \\\longrightarrow \sf x + \frac{1}{x} = \sqrt{25} \\ \\\longrightarrow \sf x + \dfrac{1}{x} = 5$

We know that

$\longrightarrow \sf{a}^3 + {b}^{3} = (a + b)( {a}^{2} + {b}^{2} -ab)$

Therefore

$\longrightarrow \sf {x}^{3} + \frac{1}{ {x}^{3}} =( x + \frac{1}{x})( {x}^{2} + \frac{1}{ {x}^{2}} - x \times \frac{1}{x} \\ \\\longrightarrow \sf {x}^{3} + \frac{1}{ {x}^{3}} = 5 \times( 23 -1)\\ \\ \longrightarrow\boxed{ \sf {x}^{3} + \dfrac{1}{ {x}^{3}} = 110}$