Question

if x=3-2√2 find x^3-1/x^3

Answer 1

Heya there !!!
Here is the answer you were looking for:

Identities used :

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


$x = 3 - 2 \sqrt{2} \\ \\ \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} }$

On rationalizing the denominator we get,

$\frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} } \\ \\ \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{( { (3)}^{2} - {( 2\sqrt{2} )}^{2} } \\ \\ \frac{1}{x} = \frac{3 + 2 \sqrt{2} }{9 - 8} \\ \\ \frac{1}{x} = 3 + 2 \sqrt{2} \\ \\ x - \frac{1}{x}$

Putting the values we get

$x - \frac{1}{x} = (3 - 2 \sqrt{2} ) - (3 + 2 \sqrt{2} ) \\ \\ x - \frac{1}{x} = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} \\ \\ x - \frac{1}{x} = 6$

Now cubing both the sides we get

${(x - \frac{1}{x} )}^{3} = {(6)}^{3} \\ \\ {(x)}^{3} - {( \frac{1}{x} )}^{3} - 3 \times x \times \frac{1}{x} (x - \frac{1}{x} ) = 216 \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } - 3(6) = 216 \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } - 18 = 216 \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } = 216 + 18 \\ \\ {x}^{3} - \frac{1}{ {x}^{3} } = 234$

Hope this helps!!!

If you have any doubt regarding to my answer, please ask in the comment section ^_^

@Mahak24

Thanks...
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