If $$X+Y+Z=6$$ and $$XY+ZX+ZY=10$$, then find the value of $$X^3+Y^3+Z^3-3XYZ$$.

Given : $$xy+yz+zx=10$$ -------------(i)

Also, $$x+y+z=6$$ ------------(ii)

Squaring both sides, we get :

=> $$(x+y+z)^2=(6)^2$$

=> $$(x^2+y^2+z^2)+2(xy+yz+zx)=36$$

Substituting value from equation (i), 

=> $$x^2+y^2+z^2+2(10)=36$$

=> $$x^2+y^2+z^2=36-20=16$$ ------------(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),

= $$(6)(16-10)$$

= $$6\times6=36$$

=> Ans - (C)