Question

If
$x + \frac{1}{x} = 6 \: \: find \: i \: \: \: \: {x}^{2} + \frac{1}{ {x}^{2} } \: \:$
$ii \: {x}^{3} + \frac{1}{ {x}^{3} }$

Answer 1

Hey !!!

x + 1 /x = 6 -----------1)

(a + b )² - 2ab = a² + b²
similarly like that .

(x + 1 /x )² - 2x×1/x = x² + 1/x²


6² - 2 = x² + 1/x²

34 = x² + 1/x² ---------2)

We know that.

a³ + b³ = (a + b) ( a² + b² - ab )

like that.

x³ + 1/x³ = (x + 1/x ) ( x² + 1/x² - x×1/x )

x³ + 1/x³ = 6 × (34 - 1 )

x³ + 1/x³ = 6×33

x³ + 1/x³ = 198 Answer !!!

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Hope it helps you !!!

@Rajukumar111
$.$

Answer 2

Heya !!

Here's your answer.. ⬇⬇
________________________________

$x + \frac{1}{x} = 6 \\ \\ {(x + \frac{1}{x} })^{2} = {(6)}^{2} \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 36 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 36 - 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 34 \\ \\ = = = = = = = = = = = = \\ \\ x + \frac{1}{x} = 6 \\ \\ {(x + \frac{1}{x} )}^{3} = {(6)}^{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3x \times \frac{1}{x} (x + \frac{1}{x} ) = 216 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(6) = 216 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 216 - 18 \\ \\ {x}^{3} + \frac{1}{ {x}^{ 3} } = 198$
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Hope it helps..
Thanks :))