Question

plz solve it plz


plz do part (a) and (f)​

Answer 1

$\red{\sf \: x = 2 - \sqrt{3} \: find \: \bigg( x - \frac{1}{x} { \bigg)}^{3} }$

$\sf \green{given}$

$\: \: \: \: \: \: \: \: \: \: \sf \: x = 2 - \sqrt{3}$

$\sf \green{to \: find :} \bigg(x - \frac{1}{ x} { \bigg)}^{3}$

$\sf \green{solution :}x = 2 - \sqrt{3}$

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

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

$\tt \: \frac{1}{x} = \frac{2 + \sqrt{3} }{4 - 3}$

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

$\sf \green{now}$

$\tt \: x - \frac{1}{x} = 2 - \sqrt{3} - 2 \sqrt{3}$

$\tt \: x - \frac{1}{x} = - 2 \sqrt{3}$

$\sf \green{and}$

$\tt \: { \bigg(x - \frac{1}{x} \bigg) }^{3}$

$\tt - \bigg(2 \sqrt{3} { \bigg)}^{2}$

$\tt \therefore \boxed{-24 \sqrt{3} }$

____________________________________

$\sf \red{x = 2 - \sqrt{3} \: find \: {x}^{3} + \frac{1}{ {x}^{3} } }$

$\sf \green{given : }x = 2 + \sqrt{3}$

$\sf \green{ to \: find : } {x}^{3} + \frac{1}{ {x}^{3} }$

$\sf \underline \green{ solution : }x = 2 \sqrt{3}$

$\tt\: \frac{1}{x} = \frac{1}{2 + \sqrt{2} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$\tt \: \frac{1}{x} = \frac{2 - \sqrt{3} }{ {(2)}^{2} - {(3)}^{2} }$

$\tt \: \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3}$

$\tt \: \frac{1}{x} = 2 - \sqrt{3}$

$\sf \green{now}$

$\: \: \: \: \: \: \: \: \: \tt \: x + \frac{1}{x}$

$\tt \implies2 + \cancel{ \sqrt{3} } + 2 - \cancel{ \sqrt{3} }$

$\tt \implies \: 2 + 2$

$\tt \implies4$

So, on cubing both sides, we have

$\sf \bigg(x + \frac{1}{x} { \bigg)}^{3} = {(4)}^{3}$

$\sf \: {x}^{3} + \frac{1}{ {x}^{3} } + 3 \bigg(x + \frac{1}{x} \bigg) = 64$

$\sf \: {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 4 = 64$

$\sf \: {x}^{3} + \frac{1}{ {x}^{3} } + 12 = 64$

$\sf \: {x}^{3} + \frac{1}{ {x}^{3} } = 64 - 12$

$\sf \: {x}^{3} + \frac{1}{ {x}^{3} } = 52$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \boxed{ \green{ \sf{x}^{3} + \frac{1}{ {x}^{3} } = 52}}$