\( \Large \frac{ \left(m-n\right)^{3}+ \left(n-r\right)^{3}+ \left(r-m\right)^{3} }{6 \left(m-n\right) \left(n-r\right) \left(r-m\right) }=? \)
Correct Answer: A) \( \Large \frac{1}{2} \) |
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Description for Correct answer:
Let, \( \Large \left(m-n\right)=a, \left(n-r\right)=b, \left(r-m\right)=c \)
we get
\( \Large a+b+c = 0 \)
Therefore, \( \Large a^{3}+b^{3}+c^{3} = 3abc \)
Therefore, Given expression \( \Large \frac{a^{3}+b^{3}+c^{3}}{6 abc} = \frac{3abc}{6abc}=\frac{1}{2} \)
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