Factor m3 + 6m2 + 5m into m(m2 + 6m + 5) = m(m+5)(m+1). We can then see that this product is always a multiple of 3. If m = 0 mod 3, then m is divisible by 3. If m = 1 mod 3, then m + 5 is a multiple of 3. If m = 2 mod 3, then m + 1 is a multiple of 3. So in any case, some factor is a multiple of 3.
On the other hand, 27n3 + 9n2 + 9n + 1 is never a multiple of 3. Each of the first three terms is always a multiple of 3, so the entire expression is always 1 mod 3.
Thus there are no integer values of m and n giving the two expressions the same value.
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