Sum of a number and its reciprocal.

Question

If $x = 2 + \sqrt{3}$ then find $x^3 + \dfrac1{x^3}$.

My attempt: we know that the sum of a number and its reciprocal is always greater than or equal to $2$ but I don't know how to find this sum.

Answer 1

$$(2+\sqrt3)^3+\frac{1}{(2+\sqrt3)^3}=(2+\sqrt3)^3+(2-\sqrt3)^3=2\cdot2^3+2\cdot3\cdot2\cdot\left(\sqrt3\right)^2=52.$$

Answer 2

Hint:

$$\begin{align}x &= 2 + \sqrt3\\x^3 &= (2 + \sqrt3)^3 =\,?_1 \\ \frac1{x^3} &= \dfrac1{2 + \sqrt3} = \frac1{2 + \sqrt3}\cdot\frac{2 - \sqrt3}{2 - \sqrt3} =\,?_2\\ x^3 + \frac1{x^3} &=\,?_1 + ?_2\end{align}$$

Also, what is $(a + b)^3$ and $(a - b)^3$? What happens when you have $(a + b)^3 + (a - b)^3$?

Answer 3

$\left(x+\frac1x\right)^3=x^3+\frac1{x^3}+3\left(x+\frac1x\right),$ so $x^3+\frac1{x^3}=(x+\frac1x)^3-3(x+\frac1x)$.

When $x=2+\sqrt3, $ $\frac1x=\frac1{2+\sqrt3}\frac{2-\sqrt3}{2-\sqrt3}=2-\sqrt3$,

so $x+\frac1x=4$, so $\left(x+\frac1x\right)^3=4^3-3(4)=52.$