If x + y = 4, xy = 2, y + z = 5, yz = 3, z + x = 6 and zx = 4, then find the value of $$x^3 + y^3 + z^3 — 3xy$$.

Given,

$$x+y=4$$ and $$xy=2$$

$$y+z=5$$ and $$yz=3$$

$$z+x=6$$ and $$zx=4$$

$$x^3 + y^3 + z^3 — 3xy$$ = $$\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)$$

= $$\frac{2}{2}\left(x+y+z\right)\frac{2}{2}\left(x^2+y^2+z^2-xy-yz-zx\right)$$

= $$\frac{1}{2}\left(2x+2y+2z\right)\frac{1}{2}\left(2x^2+2y^2+2z^2-2xy-2yz-2zx\right)$$

= $$\frac{\left(x+y\right)+\left(y+z\right)+\left(z+x\right)}{2}.\frac{\left(x^2+y^2+2xy+y^2+z^2+2yz+z^2+x^2+2zx-4xy-4yz-4zx\right)}{2}$$

= $$\frac{\left(4\right)+\left(5\right)+\left(6\right)}{2}.\frac{\left(\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2-4\left(2\right)-4\left(3\right)-4\left(4\right)\right)}{2}$$

= $$\frac{15}{2}.\frac{\left(4^2+5^2+6^2-8-12-16\right)}{2}$$

= $$\frac{15}{2}.\frac{\left(16+25+36-36\right)}{2}$$

= $$\frac{15}{2}.\frac{41}{2}$$

= $$153.75$$

Hence, the correct answer is Option C