Find the number of triples of integers a<b<c with a≥−10, such that a(b−c)^3+b(c−a)^3+c(a−b)^3=0
To find the number of triples of integers that satisfy the given equation, we can analyze each possible value of a, b, and c within the given range.
Let's start with a = -10. Substituting a = -10 into the equation, we have:
-10(b - c)^3 + b(c - (-10))^3 + c((-10) - b)^3 = 0
To simplify the equation, let's consider the terms containing b and c:
b(c - (-10))^3 + c((-10) - b)^3
Expand the cube terms:
b(c + 10)^3 + c(-10 - b)^3
Simplify further:
b(c + 10)(c + 10)(c + 10) + c(-10 - b)(-10 - b)(-10 - b)
Since we are looking for integer solutions, b and c must have the same parity (both even or both odd). By considering the expression above, we see that if b and c have the same parity, the entire expression will be even.
Now, let's analyze the possible values of b and c:
- If b and c are both odd, the expression will be even.
- If b and c are both even, the expression will be even.
Therefore, for a = -10, there are no restrictions on the possible values of b and c.
Moving on to a = -9, we substitute a = -9 into the equation and simplify as we did before. We continue this process for each value of a from -9 to -1. In each case, we find that there are no restrictions on the possible values of b and c.
Finally, for a = 0 to a = 9, we perform the same steps, and in each case, we find that there are no restrictions on the possible values of b and c.
Thus, for each value of a from -10 to 9, there are no restrictions on the possible values of b and c. Therefore, the number of triples of integers (a, b, c) that satisfy the given equation is infinity.