Simplify : $$\frac{{{{1.5}^3} + {{4.7}^3} + {{3.8}^3} - 3 \times 1.5 \times 4.7 \times 3.8}}{{{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}}$$ = ?
A. 0
B. 1
C. 10
D. 30
Answer: Option C
Solution(By Examveda Team)
$$\frac{{{{1.5}^3} + {{4.7}^3} + {{3.8}^3} - 3 \times 1.5 \times 4.7 \times 3.8}}{{{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}}$$$$ = \frac{{\left( {1.5 + 4.7 + 3.8} \right)\left\{ {{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}{{\left\{ {{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}$$
$$\left[ {\therefore {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)} \right]$$
$$\eqalign{ & = 1.5 + 4.7 + 3.8 \cr & = 10.0 \cr & = 10 \cr} $$
Related Questions on Surds and Indices
A. $$\frac{1}{2}$$
B. 1
C. 2
D. $$\frac{7}{2}$$
Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
A. 1.45
B. 1.88
C. 2.9
D. 3.7
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