| Simplify \[\frac{\begin{align} & 24\times 24\times 24+32\times 32\times 32+45\times 45 \\ & \times \,\,45-3\times 24\times 32\times 45 \\ \end{align}}{\begin{align} & 24\times 24\times 32\times 32+45\times 45-24\times 32 \\ & -\,\,32\times 45-45\times 24 \\ \end{align}}\] |
A) 105
B) 101
C) 103
D) 107
Correct Answer: B
Solution :
| Given expression is of the form |
| \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] |
| \[=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca)\] |
| \[\therefore \]\[\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca}=a+b+c\] |
| \[\therefore \]\[a+b+c=24+32+45=101\] |
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