Question

if x = 2 - √3, find the value of x³ -1/x³​

Answer 1

Answer:

-30√3

Step-by-step explanation:

x=2-√3 1/x =2+√3

x²=7-4√3

1/x²=7+4√3

x³-1/x³ in form a³-b³ = (a-b)(a²+b²+ab)

= (x-1/x)(x²+1/x²+x.1/x)

=-2√3(14+1)=-30√3

Answer 2

GIVEN :-

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

TO FIND :-

$\\ \sf \: {x}^{3} - \dfrac{1}{ {x}^{3} }$

SOLUTION :-

We know ,

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

Taking reciprocal ,

$\\ \sf \: \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3} }$

Rationalising the denominator ,

Rationalising factor is 2+√3

$\\ \sf \: \dfrac{1}{x} = \dfrac{1}{2 - \sqrt{3} } \times \dfrac{2 + \sqrt{3} }{2 + \sqrt{3} } \\ \\ \\ \sf \: \dfrac{1}{x} = \dfrac{2 + \sqrt{3} }{(2 - \sqrt{3})(2 + \sqrt{3}) } \\ \\ \\ \boxed{\sf \: \dfrac{1}{x} = 2 + \sqrt{3} }$

We have ,

• x = 2 - √3
• 1/x = 2 + √3

So,

$\\ \sf \: x - \frac{1}{x} = 2 - \sqrt{3} - (2 + \sqrt{3} ) \\ \\ \\ \sf \: x - \frac{1}{x} = \cancel2 - \sqrt{3} - \cancel2 - \sqrt{3} \\ \\ \\ \sf \: x - \frac{1}{x} = - 2 \sqrt{3}$

Cubing both sides , we get..

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

We know , (a - b)³ = a³ - 3a²b + 3ab² - b³

Here ,

• a = x
• b = 1/x

Putting values we get...

$\\ \sf \: {x}^{3} - 3 {x}^{2} \left( \frac{1}{x} \right) + 3x {\left( \frac{1}{x} \right)}^{2} - {\left( \frac{1}{x} \right) }^{3} = - 24 \sqrt{3} \\ \\ \\ \sf \: {x}^{3} - 3x + \frac{3}{x} - \frac{1}{ {x}^{3} } = - 24 \sqrt{3} \\ \\ \\ \sf \: {x}^{3} - \frac{1}{ {x}^{3} } - 3x + \frac{3}{x} = - 24 \sqrt{3} \\ \\ \\ \sf \: {x}^{3} - \frac{1}{ {x}^{3} } - 3\left(x + \frac{1}{x} \right)= - 24 \sqrt{3}$

$\\ \sf \: {x}^{3} - \frac{1}{ {x}^{3} } - 3 \{2 - \cancel{\sqrt{3}} + 2 + \cancel{\sqrt{3}} \} = - 24 \sqrt{3} \\ \\ \\ \sf \: {x}^{3} - \frac{1}{ {x}^{3} } - 3(4) = - 24 \sqrt{3} \\ \\ \\ \sf \: {x}^{3} - \frac{1}{ {x}^{3} } = - 24 \sqrt{3} + 12 \\ \\ \\ \sf \: {x}^{3} - \frac{1}{ {x}^{3} } = - 24 \sqrt{3} + 4 \sqrt{3} \\ \\ \\ \boxed{ \bf \: {x}^{3} - \frac{1}{ {x}^{3} } = - 20 \sqrt{3} }$