If x = 32.5, y = 34.6 and z = 30.9, then the value x³ + y³ + z³ - 3xyz of is 0.98k, where k is equal to:

Solution (By Examveda Team)

$$\eqalign{ & x = 32.5,\,y = 34.6{\text{ and }}z = 30.9 \cr & {x^3} + {y^3} + {z^3} - 3xyz \cr & = \left( {x + y + z} \right)\left[ {\frac{1}{2}\left\{ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right\}} \right] \cr & = \left( {32.5 + 34.6 + 30.9} \right)\left[ {\frac{1}{2}\left\{ {{{\left( {32.5 - 34.6} \right)}^2} + {{\left( {34.6 - 30.9} \right)}^2} + {{\left( {30.9 - 32.5} \right)}^2}} \right\}} \right] \cr & = 98\left[ {\frac{1}{2}\left\{ {{{\left( { - 2.1} \right)}^2} + {{\left( {3.7} \right)}^2} + {{\left( { - 1.6} \right)}^2}} \right\}} \right] \cr & = 98\left[ {\frac{1}{2}\left\{ {4.41 + 13.69 + 2.56} \right\}} \right] \cr & = 98\left[ {\frac{1}{2}\left\{ {20.66} \right\}} \right] \cr & = 98 \times 10.33 \cr & = 1012.34 \cr & {x^3} + {y^3} + {z^3} - 3xyz = 0.98k \cr & 1012.34 = 0.98k \cr & k = \frac{{1012.34}}{{0.98}} = 1033 \cr} $$