Determine the value of $$(\frac{1}{r}+\frac{1}{s})$$ when $$r^{3}+s^{3}=0$$ and $$r+s=6$$

Given : $$r^{3}+s^{3}=0$$ and $$r+s=6$$ ---------(i)

$$\because$$ $$x^3+y^3=(x+y)(x^2+y^2-xy)$$

=> $$(r+s)(r^2+s^2-rs)=0$$

=> $$6(r^2+s^2-rs)=0$$

=> $$r^2+s^2-rs=0$$

=> $$r^2+s^2=rs$$ -------------(ii)

Also, squaring equation (i),

=> $$(r+s)^2=(6)^2$$

=> $$r^2+s^2+2rs=36$$

Substituting value from equation (ii), => $$rs+2rs=3rs=36$$

=> $$rs=\frac{36}{3}=12$$ ------------(iii)

$$\therefore$$ $$(\frac{1}{r}+\frac{1}{s})=\frac{r+s}{rs}$$

Dividing equation (i) by (iii),

= $$\frac{6}{12}=0.5$$

=> Ans - (B)