Algebraic Manipulation

$\begin{aligned} r^3+ \dfrac{1}{r^3} & = \left ( r+ \dfrac{1}{r} \right ) \left (r^2- r \cdot \dfrac{1}{r} + \dfrac{1}{r^2} \right ) \\ & = \left ( r+ \dfrac{1}{r} \right ) \left( r^2 + \dfrac{1}{r^2}-1 \right ) \\ & = \left ( r+ \dfrac{1}{r} \right ) \left [ \left (r+ \dfrac{1}{r} \right )^2 -2-1 \right ] \\ & = \left ( r+ \dfrac{1}{r} \right ) \left ( 3-2-1 \right ) \\ &= 0 \\ \end{aligned}$

from given equation , we get r + 1/r = root(3)...........so using ( r+1/r )^3 = r^3 +1/r^3 +3 r 1/r*( r+1/r)....... by putting value of r+1/r in above formula....we get r^3 + 1/r^3 = 0

$\left( r+\dfrac {1}{r} \right)^2 = 3\quad \Rightarrow r^2 + 2 + \dfrac {1}{r^2} = 3\quad \Rightarrow r^2 + \dfrac {1}{r^2} = 1$

Now, we have:

$r^3 + \dfrac {1}{r^3} = \left( r + \dfrac {1}{r} \right) \left( r^2 + \dfrac {1}{r^2} - 1\right) = \left( r + \dfrac {1}{r} \right) \left( 1 - 1 \right)$

$\quad \quad \quad \quad = \left( r + \dfrac {1}{r} \right) \left( 0 \right) = \boxed{0}$

r + 1/r = \sqrt{3}

r^2 + 2 + 1/r^2 = 3

r^2 + 1/r^2 = 1

r^3 + 1/ r^3 = (r + 1/r)(r^2 + 1/r^2 -1)

(\sqrt{3})(1-1)

(\sqrt{3})(0) = 0

the given question is in the form of (a+b)^3