| \[\frac{\begin{align} & {{(2.247)}^{3}}+{{(1.730)}^{3}}+{{(1.023)}^{3}} \\ & -\,\,3\times 2.247\times 1.730\times 1.023 \\ \end{align}}{\begin{align} & {{(2.247)}^{2}}+{{(1.730)}^{2}}+{{(1.023)}^{2}}-2.247 \\ & \times 1.730-1.730\times 1.023-2.247\times 1.023 \\ \end{align}}=?\] |
A) 1.730
B) 4
C) 5
D) 5.247
Correct Answer: C
Solution :
| We know that, \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] |
| \[=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)\] |
| \[\Rightarrow \]\[(a+b+c)=\left( \frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca} \right)\] |
| (i) |
| Given that, |
| \[=\frac{{{(2.47)}^{3}}+{{(1.730)}^{3}}+{{(1.023)}^{3}}-3\times 2.47\times 1.730\times 1.023}{\begin{align} & {{(2.247)}^{2}}+{{(1.730)}^{2}}+{{(1.023)}^{2}}-(2.247)\times (1.730) \\ & -\,\,(1.730\times 1.023)-(2.247)\times 1.023) \\ \end{align}}\]\[=(2.247+1.730+1.023)\] [from Eq.(i)] |
| \[=5.000=5\] |
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