A) 0
B) 1
C) 10
D) 30
Correct Answer: C
Solution :
\[\frac{{{(1.5)}^{3}}+{{(4.7)}^{3}}+{{(3.8)}^{3}}-3\times 1.5\times 4.7\times 3.8}{{{(1.5)}^{2}}+{{(4.7)}^{2}}+{{(3.8)}^{2}}-1.5\times 4.7-4.7\times 3.8-3.8\times 1.5}\] \[=1.5+4.7+3.8=10\] \[\left[ \because \frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-ab-ca}=a+b+c \right]\]You need to login to perform this action.
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