Question

please solve this guy's immediately....

Answer 1

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

Identity used :

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

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

On rationalizing the denominator we get,

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

1. On squaring both the sides (x) + (1/x) = (2√3)

${(x + \frac{1}{x} )}^{2} = {(2 \sqrt{3} )}^{2} \\ \\ {(x)}^{2} + {( \frac{1}{x} )}^{2} + 2 \times x \times \frac{1}{x} = 12 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 12 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 12 - 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 10$

2. On cubing both the sides of (x) + (1/x) = (2√3)

${(x + \frac{1}{x}) }^{3} = {(2 \sqrt{3} )}^{2} \\ \\ {(x)}^{3} + {( \frac{1}{x} )}^{3} + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) = 24 \sqrt{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3(2 \sqrt{3} ) = 24 \sqrt{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 6 \sqrt{3} = 24 \sqrt{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 24 \sqrt{3} - 6 \sqrt{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 18 \sqrt{3}$

3. On squaring both the sides of (x)^2 + (1/x)^2 = 10

${( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = {(10)}^{2} \\ \\ {( {x}^{2} )}^{2} + {( \frac{1}{ {x}^{2} } )}^{2} + 2 \times {x}^{2} \times \frac{1}{ {x}^{2} } = 100 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 100 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } = 100 - 2 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } = 98$

Hope you understood !!!

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

@Mahak24

Thanks...
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Answer 2

Hey sistah!!!

Here's your answer =>=>

Refer to the attached file ^^^^

Sorry if you feel some difficulty in understanding the solution, but plz ask me. I'll try to explain you.

Hope it will help you ☆▪☆

Thanks ^_^