kaliprasad said:Show that for integer a,b,c the product $abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$ is divisible by 7
To prove that the product is divisible by 7, we can use the fact that 7 is a prime number. This means that any number that is not divisible by 7 can be expressed as a product of other numbers that are also not divisible by 7. Therefore, if we can show that each term in the product is divisible by 7, we can conclude that the whole product is divisible by 7.
The key properties of integers that make this product divisible by 7 are that they can be expressed as whole numbers and they can be multiplied, divided, and added or subtracted from one another. This allows us to manipulate the terms in the product to show that they are all divisible by 7.
For example, if we take a = 3, b = 4, and c = 5, we get the product 3*4*5*(3^3-4^3)*(4^3-5^3)*(5^3-3^3) = 60*35*(-91)*(-152)*(-152). We can see that each term is divisible by 7, and therefore the whole product is divisible by 7.
Yes, this product is also divisible by 2 and 3 for all integer values of a, b, and c. This is because 7, 2, and 3 are all relatively prime, meaning they have no common factors other than 1. Therefore, any integer that is divisible by all three of these numbers must also be divisible by their product (7*2*3 = 42).
Yes, this proof can be extended to show that the product is divisible by any prime number, as long as the integer values of a, b, and c are relatively prime to that prime number. This is because the same logic applies - if all the terms in the product are divisible by a prime number, then the whole product must also be divisible by that prime number.