If $$(1/x) + (1/y) + (1/z) = 0$$ and $$x + y + z = 11$$, then what is the value of $$x^{3}+y^{3}+z^{3}-3xyz$$ ?

Given : $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$$

=> $$\frac{yz+zx+xy}{xyz}=0$$

=> $$xy+yz+zx=0$$ -------------(i)

Also, $$x+y+z=11$$ ------------(ii)

Squaring both sides, we get :

=> $$(x+y+z)^2=(11)^2$$

=> $$(x^2+y^2+z^2)+2(xy+yz+zx)=121$$

Substituting value from equation (i), 

=> $$x^2+y^2+z^2=121$$ ------------(iii)

To find : $$x^{3}+y^{3}+z^{3}-3xyz$$ 

= $$(x+y+z)[(x^2+y^2+z^2)-(xy+yz+zx)]$$

Substituting values from equations (i), (ii) and (iii),

= $$(11)(121-0)$$

= $$11\times121=1331$$

=> Ans - (A)