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If \mathrm{a, b, c} are three natural numbers such that \mathrm{a+b+c}  is divisible by 6 , then \mathrm{a^{3}+b^{3}+c^{3}} must be divisible by

Option: 1

6


Option: 2

12


Option: 3

24


Option: 4

none of these


Answers (1)

best_answer

We have \mathrm{\left(a^{3}-a\right)+\left(b^{3}-b\right)+\left(c^{3}-c\right)}

\mathrm{=(a-1)(a)(a+1)+(b-1)(b)(b+1)+(c-1)(c)(c+1)}

Since \mathrm{(x-1) x(x+1)} is divisible both by 2 and 3 , it is divisible by 6 . Thus,

\begin{array}{ll} & 6 \mid\left\{\left(a^{3}-a\right)+\left(b^{3}-b\right)+\left(c^{3}-c\right)\right\} \\ \Rightarrow & 6 \mid\left(a^{3}+b^{3}+c^{3}\right) \text { as } 6 \mid(a+b+c) . \end{array}

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avinash.dongre

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