If $x+{1\over x} = r $ then what is $x^3+{1\over x^3}$? If $$x+{1\over x} = r $$ then what is $$x^3+{1\over x^3}$$

Options:
$(a) 3,$
$(b) 3r,$
$(c)r,$
$(d) 0$

$\displaystyle r^3=\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3(x)\frac{1}{x}\left(x+\frac{1}{x}\right)=x^3+\frac{1}{x^3}+3r$
$\displaystyle \Rightarrow r^3-3r=x^3+\frac{1}{x^3}$
Your options are incorrect.For a quick counter eg. you can take $x=1/2$ to get $r=\frac{5}{2}$ and $x^3+\frac{1}{x^3}=\frac{65}{8}$ but none of the options result in $65/8$
