$\begin{aligned} Simplify: \\ \frac{{(1.5)}^{3} + {(4.7)}^{3} + {(3.8)}^{3} - 3 \times 1.5 \times 4.7 \times 3.8}{{(1.5)}^{2} + {(4.7)}^{2} + {(3.8)}^{2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} = ? \end{aligned}$ $\eqalign{ & {\text{Simplify:}} \cr & \frac{{{{\left( {1.5} \right)}^3} + {{\left( {4.7} \right)}^3} + {{\left( {3.8} \right)}^3} - 3 \times 1.5 \times 4.7 \times 3.8}}{{{{\left( {1.5} \right)}^2} + {{\left( {4.7} \right)}^2} + {{\left( {3.8} \right)}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}} = ? \cr}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{{(1.5)}^{3} + {(4.7)}^{3} + {(3.8)}^{3} - 3 \times 1.5 \times 4.7 \times 3.8}{{(1.5)}^{2} + {(4.7)}^{2} + {(3.8)}^{2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \\ = \frac{{{(1.5)}^{2} + {(4.7)}^{2} + {(3.8)}^{2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}}{{{(1.5)}^{2} + {(4.7)}^{2} + {(3.8)}^{2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}} \\ . . . . [∴ a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)] \\ = 1.5 + 4.7 + 3.8 \\ = 10.0 \\ = 10 \end{aligned}

$\eqalign{ & \frac{{{{\left( {1.5} \right)}^3} + {{\left( {4.7} \right)}^3} + {{\left( {3.8} \right)}^3} - 3 \times 1.5 \times 4.7 \times 3.8}}{{{{\left( {1.5} \right)}^2} + {{\left( {4.7} \right)}^2} + {{\left( {3.8} \right)}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}} \cr & = \frac{{\left\{ {{{\left( {1.5} \right)}^2} + {{\left( {4.7} \right)}^2} + {{\left( {3.8} \right)}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}{{\left\{ {{{\left( {1.5} \right)}^2} + {{\left( {4.7} \right)}^2} + {{\left( {3.8} \right)}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}} \cr & ....\left[ {\therefore {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)} \right] \cr & = 1.5 + 4.7 + 3.8 \cr & = 10.0 \cr & = 10 \cr}$