Advertisements
Advertisements
Question
If x = \[\sqrt[3]{2 + \sqrt{3}}\] , then \[x^3 + \frac{1}{x^3} =\]
Options
2
4
8
9
Advertisements
Solution
Given that . `x = 3sqrt(2+sqrt3)` It can be simplified as
` x^3 = 2+sqrt3`
`1/ x^3 = 1 /(2+sqrt3)`
We know that rationalization factor for `2+sqrt3` is `2- sqrt3`. We will multiply numerator and denominator of the given expression `1/(2+sqrt3)`by `2-sqrt3`, to get
`1/x^3 = 1/(2+sqrt3 ) xx (2-sqrt3)/(2-sqrt3)`
`= (2-sqrt3)/((2)^2 - (sqrt3)^2)`
`= (2-sqrt3)/(4-3)`
`=2-sqrt3`
Therefore,
`x^3 + 1/x^3 = 2 +sqrt3 +2 - sqrt3`
`= 2+2`
`=4`
APPEARS IN
RELATED QUESTIONS
Prove that:
`sqrt(1/4)+(0.01)^(-1/2)-(27)^(2/3)=3/2`
Show that:
`1/(1+x^(a-b))+1/(1+x^(b-a))=1`
Show that:
`{(x^(a-a^-1))^(1/(a-1))}^(a/(a+1))=x`
Find the value of x in the following:
`(2^3)^4=(2^2)^x`
Find the value of x in the following:
`(13)^(sqrtx)=4^4-3^4-6`
For any positive real number x, find the value of \[\left( \frac{x^a}{x^b} \right)^{a + b} \times \left( \frac{x^b}{x^c} \right)^{b + c} \times \left( \frac{x^c}{x^a} \right)^{c + a}\].
If (23)2 = 4x, then 3x =
Which one of the following is not equal to \[\left( \frac{100}{9} \right)^{- 3/2}\]?
If g = `t^(2/3) + 4t^(-1/2)`, what is the value of g when t = 64?
The value of \[\frac{\sqrt{48} + \sqrt{32}}{\sqrt{27} + \sqrt{18}}\] is
