Question

anyone solve this as soon as possible
it's quite urgent... ​

Answer 1

Given,

${ \huge{ \boxed{x + \frac{1}{x} = 4}}}$

Now,

${(x + \frac{1}{x} )}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = { \boxed{ {x}^{2} + \frac{1}{ {x}^{2} } = 14 }} \\ \\ {(x - \frac{1}{x} )}^{2} = > {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ {(x - \frac{1}{x} )}^{2}= > 14 - 2 = 12 \\ \\ \: \: \: \: \: \: = > { \boxed{ \large{ \red{x - \frac{1}{x} = \sqrt{12} }}}}$

Again,

${x}^{3} + \frac{1}{ {x}^{3} } = >(x + \frac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } - 1 ) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > (4)(14 - 1) \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > 56 - 4 = { \red{52}}$

Thus,

${x}^{3} - \frac{1}{ {x}^{3} } = > (x - \frac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } + 1 )\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > ( \sqrt{12} )(14 + 1)\\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > 2 \sqrt{3} (15)\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > { \red{30 \sqrt{3} = \sqrt{2700} }}$