Question

If x = 3 + 2√2 , then find the value of x² + 1 / x² and x³ + 1 / x³.​

Answer 1

$\huge{✪}$$\huge{\underline{\mathcal{Answer}}}$

1. 34
2. 198

$\huge{\underline{\mathcal{Formulas\:Used}}}$

• $x^2+\dfrac{1}{x^2}={(x+\dfrac{1}{x})}^2-2$
• $x^3+\dfrac{1}{x^3}=(x+\dfrac{1}{x})(x^2+\dfrac{1}{x^2}-1)$

$\huge{\underline{\mathcal{Steps:}}}$

$\dfrac{1}{x} = \dfrac{1}{3 + 2 \sqrt{2} } \\ \\ \dfrac{1}{x} = \dfrac{1}{3 + 2 \sqrt{2} } \times \dfrac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \dfrac{1}{x} = 3 - 2 \sqrt{2}$

Now adding x and $\dfrac{1}{x}$ to get $x+\dfrac{1}{x}$

$x + \dfrac{1}{x} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ x + \dfrac{1}{x} = 6 \: \: \: ....(1)$

Part 1:

${x}^{2} + \dfrac{1}{ {x}^{2} } = {(x + \dfrac{1}{x} )}^{2} - 2 \\ \\ {6}^{2} - 2 \\ \\ 36 - 2 \\ \\ 34$

Part 2:

${x}^{3} + \dfrac{1}{ {x}^{3} } = (x + \dfrac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } - 1) \\ \\ {x}^{3} + \dfrac{1}{ {x}^{3} } = 6(34 - 1) \\ \\ {x}^{3} + \dfrac{1}{ {x}^{3} } = 6 \times 33 = 198$

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Answer 2

Answer:

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