Question

plz someone help me to do this no.​

Answer 1

Answer:

your answer attached in the photo

Answer 2

Answer:

a) 0

b) ±√6

c) 3√6

Step-by-step explanation:

Given:

$x - \dfrac{1}{x} = \sqrt{2}$

To find:

$a) {x}^{2} + \dfrac{1}{ {x}^{2} }$

$b)x + \dfrac{1}{x}$

$c){x}^{3} + \dfrac{1}{ {x}^{3} }$

Solution:

$a) x - \dfrac{1}{x} = \sqrt{2}$

[Squaring both side ]

${(x - \frac{1}{x} )}^{2} = {( \sqrt{2} )}^{2} \\ \\ {x}^{2} + \dfrac{1}{ {x}^{2} } - 2.x. \frac{1}{x} = 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } - 2 = 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 2 + 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 4$

Hence, The required value of a is 0

$b)x {}^{2} + \dfrac{1}{x {}^{2} } = 4 \: \: \: \: \: (from \: \: \: \: a) \\ \\ {x}^{2} + 2 + \frac{1}{ {x}^{2} } = 6\\ \\ {x}^{2} + 2.x. \frac{1}{x} + \frac{1}{ {x}^{2} } = 6 \\ \\ {(x + \frac{1}{x}) }^{2} = 6 \\ \\ x + \frac{1}{x} = \pm \sqrt{6}$

Hence the required value of b is ± √6

$c)x + \dfrac{1}{x} = \sqrt{6} \: \: \: \: \: \: \: (from \: \: \: b)$

[cubing both side ]

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

Hence the required value of c is 3√6