A better solution would be choose any prime \[p\] dividing \[a,b,c\] and consider the maximum powers of \[p\] in \[a,b,c\] respectively \[\alpha,\beta,\gamma\] and WLOG let \[\alpha\ge \beta \ge \gamma\] .Now the max power of \[p\] dividing \[(a+b+c)^{31}\] is \[p^{31\gamma}\] .It suffices to show that \[\alpha+\beta+\gamma\le 31\gamma\] but \[\alpha\le 5\beta\le 25\gamma\] from the given condition.So summing them we get the desired inequality.This helps an easy generalization:
let \[a,b,c\] be natural numbers such that \[a\mid b^n,b\mid c^n,c\mid a^n\] then \[abc\mid (a+b+c)^{n^2+n+1}\]

Well, first of all we note that a divides \(c^{25}\): if b divides \(c^5\), then \(b^5\) divides \(c^{25}\). Because a divides \(b^5\), it divides also \(c^{25}\). This can be applied cyclically: b divides \(a^{25}\) and c divides \(b^{25}\).

In the factorization of \( (a+b+c+)^{31} \) there will be three members \(a^{31};b^{31};c^{31}\), members like \(n\cdot a^k \cdot b^{31-k}\) (and cyclical, i.e. others with ac or bc instead of ab) and members like \(abc \cdot something\). Clearly, \(abc\) divides all the members of the third type. We can express \(a^{31}\) as \(a^{25}\cdot a^5 \cdot a\), for the observation written above, it's dividible by \(abc\).

Remains the second case:let's assume WLOG that's with the letters a and b.

We have three cases: k<5,k=5 and k>5.

In the first one, \(31-k\geq 26\), so we can express \(a^k\cdot b^{31-k}\) as \(a^k cdot b^{25}\cdot b^h\) where h is a positive integer. a divides \(a^k\), b divides \(b^h\) and c divides \(b^25\) so abc divides \(a^k\cdot b^{31-k}\) .

In the second case,\(a^k\cdot b^{31-k}\) is \(a^5\cdot b^{21}\cdot b^5\). c divides \(a^5\), b divides \(b^{21}\) and a divides \(b^5\).

In the third case, we can express \(a^k\cdot b^{31-k}\) as \(a^5 cdot a^j \cdot b^{31-k} \) where j is a positive integer. c divides \(a^5\), b divides \(b^{31-k}\) and a divides \(a^j\). Q.E.D.

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TopNewestA better solution would be choose any prime \[p\] dividing \[a,b,c\] and consider the maximum powers of \[p\] in \[a,b,c\] respectively \[\alpha,\beta,\gamma\] and WLOG let \[\alpha\ge \beta \ge \gamma\] .Now the max power of \[p\] dividing \[(a+b+c)^{31}\] is \[p^{31\gamma}\] .It suffices to show that \[\alpha+\beta+\gamma\le 31\gamma\] but \[\alpha\le 5\beta\le 25\gamma\] from the given condition.So summing them we get the desired inequality.This helps an easy generalization: let \[a,b,c\] be natural numbers such that \[a\mid b^n,b\mid c^n,c\mid a^n\] then \[abc\mid (a+b+c)^{n^2+n+1}\]

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Well, first of all we note that a divides \(c^{25}\): if b divides \(c^5\), then \(b^5\) divides \(c^{25}\). Because a divides \(b^5\), it divides also \(c^{25}\). This can be applied cyclically: b divides \(a^{25}\) and c divides \(b^{25}\).

In the factorization of \( (a+b+c+)^{31} \) there will be three members \(a^{31};b^{31};c^{31}\), members like \(n\cdot a^k \cdot b^{31-k}\) (and cyclical, i.e. others with ac or bc instead of ab) and members like \(abc \cdot something\). Clearly, \(abc\) divides all the members of the third type. We can express \(a^{31}\) as \(a^{25}\cdot a^5 \cdot a\), for the observation written above, it's dividible by \(abc\).

Remains the second case:let's assume WLOG that's with the letters a and b.

We have three cases: k<5,k=5 and k>5.

In the first one, \(31-k\geq 26\), so we can express \(a^k\cdot b^{31-k}\) as \(a^k cdot b^{25}\cdot b^h\) where h is a positive integer. a divides \(a^k\), b divides \(b^h\) and c divides \(b^25\) so abc divides \(a^k\cdot b^{31-k}\) .

In the second case,\(a^k\cdot b^{31-k}\) is \(a^5\cdot b^{21}\cdot b^5\). c divides \(a^5\), b divides \(b^{21}\) and a divides \(b^5\).

In the third case, we can express \(a^k\cdot b^{31-k}\) as \(a^5 cdot a^j \cdot b^{31-k} \) where j is a positive integer. c divides \(a^5\), b divides \(b^{31-k}\) and a divides \(a^j\). Q.E.D.

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