If x + y = 4, xy = 2, y + z = 5, yz = 3, z + x = 6 and zx = 4, then find the value of x³ + y³ + z³ - 3xyz.

Solution (By Examveda Team)

$$\eqalign{ & x + y = 4,\,y + z = 5,\,z + x = 6 \cr & {\text{So, }}x + y + z = \frac{{15}}{2} \cr & {\left( {x - y} \right)^2} = {\left( {x + y} \right)^2} - 4xy \cr & = {4^2} - 4 \times 2 \cr & = 8 \cr & {\left( {y - z} \right)^2} = {\left( {y + z} \right)^2} - 4yz \cr & = {5^2} - 4 \times 3 \cr & = 13 \cr & {\left( {z - x} \right)^2} = {\left( {z + x} \right)^2} - 4zx \cr & = {6^2} - 4 \times 4 \cr & = 20 \cr & {x^3} + {y^3} + {z^3} - 3xyz \cr & = \frac{{\left( {x + y + z} \right)}}{2}\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right] \cr & = \frac{1}{2} \times \frac{{15}}{2}\left[ {8 + 13 + 20} \right] \cr & = \frac{{15 \times 41}}{4} \cr & = 153.75 \cr} $$