Question

x^3-1/x^3 if x^2+1/x^2=18​

Answer 1

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Answer 2

Step-by-step explanation:

Here,

${x}^{2} + \frac{1}{ {x}^{2} } = 18$

${x}^{2} - 2x \frac{1}{x} + \frac{1}{ {x}^{2} } = 18 - 2$

$(x - \frac{1}{x} ) = \sqrt{16}$

now...

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

putting the value....

$( {x}^{3} - \frac{1}{ {x}^{3} } ) = 16 \sqrt{16} + 3 \sqrt{16} = 19 \sqrt{16}$

I am done.