Question

let x=2+root3,find the value of x cube+1÷xcube​

Answer 1

Answer:

52

Step-by-step explanation:

First, note that

$\displaystyle\left(x+\frac1x\right)^3=x^3+3x^2\bigl(\tfrac1x\bigr)+3x\bigl(\tfrac1{x^2}\bigr)+\frac1{x^3}=x^3+\frac1{x^3}+3\left(x+\frac1x\right)\\\\\Rightarrow x^3+\frac1{x^3} = \left(x+\frac1x\right)^3-3\left(x+\frac1x\right)$

So we'll aim first for finding the value of x + 1/x.

$\displaystyle x+\frac1x=2+\sqrt3+\frac{1}{2+\sqrt3}\\\\{}\qquad=2+\sqrt3+\frac{2-\sqrt3}{(2+\sqrt3)(2-\sqrt3)}\\\\{}\qquad=2+\sqrt3+\frac{2-\sqrt3}{4-3}\\\\{}\qquad=2+\sqrt3+2-\sqrt3\\\\{}\qquad=4$

It now follows that

$\displaystyle x^3+\frac1{x^3} = \left(x+\frac1x\right)^3-3\left(x+\frac1x\right)=4^3-3(4)=64-12=52$