Let a,b,c be positive integers such that a divides b^2 , b divides c^2 and c divides a^2 . Prove that abc divides (a + b + c)7 .

To prove that abc divides (a + b + c)^7, we'll need to show that (a + b + c)^7 is divisible by abc.

First, let's rewrite (a + b + c)^7 using the binomial theorem:

(a + b + c)^7 = C(7, 0)a^7 + C(7, 1)a^6b + C(7, 1)a^6c + C(7, 2)a^5b^2 + C(7, 2)a^5bc + C(7, 2)a^5c^2 + C(7, 3)a^4b^3 + C(7, 3)a^4b^2c + C(7, 3)a^4bc^2 + C(7, 3)a^4c^3 + ... + C(7, 7)b^7 + C(7, 7)b^6c + C(7, 7)b^5c^2 + C(7, 7)b^4c^3 + ... + C(7, 7)c^7

Now, let's focus on analyzing the terms of (a + b + c)^7 to determine if they are divisible by abc.

For a term to be divisible by abc, each variable (a, b, and c) must appear at least once in the term. In other words, for each term, we need to have an exponent greater than or equal to 1 for a, b, and c.

Looking at the terms, we see that for each binomial coefficient C(7, k) (where k ranges from 0 to 7), the powers of a, b, and c in that term are:

a: 7, 6, 6, 5, 5, 5, 4, 4, 4, ..., 1, 1, 1
b: 0, 1, 0, 2, 1, 0, 3, 2, 1, ..., 6, 5, 4
c: 0, 0, 1, 0, 1, 2, 0, 1, 2, ..., 4, 5, 6

Notice that in every term, the powers of a, b, and c can be paired up such that their sum is always at least 1. This is because a divides b^2, b divides c^2, and c divides a^2.

For example, in the term C(7, 1)a^6b, a has power 6, b has power 1, and c has power 0. Since a divides b^2, we can pair up b with 2 of the powers of a, leaving at least one power of a.

In general, for every term, we can pair up the powers of a, b, and c such that each variable's power is at least 1. Thus, every term in (a + b + c)^7 is divisible by abc.

Therefore, we conclude that abc divides (a + b + c)^7.

To prove that abc divides (a + b + c)^7, we need to show that (a + b + c)^7 is divisible by abc.

Given that a divides b^2, we can express b^2 = ak, where k is an integer.
Similarly, b divides c^2, so we can express c^2 = bl, where l is an integer.
Lastly, c divides a^2, so we can express a^2 = cm, where m is an integer.

Now, let's consider the expression (a + b + c)^7. We can expand this using the binomial theorem:

(a + b + c)^7 = C(7, 0)a^7 + C(7, 1)a^6b + C(7, 2)a^5b^2 + ... + C(7, 6)ab^6 + C(7, 7)b^7 + ...

Notice that any term in the expansion will contain at least one power of a, one power of b, and one power of c.

Let's focus on the terms that contain a^2, b^2, and c^2. These terms are:
C(7, 2)a^5b^2, C(7, 2)a^2bc^4, and C(7, 2)ab^4c^2, respectively.

Since we have expressed a^2, b^2, and c^2 in terms of a, b, and c, we substitute them in:

C(7, 2)(cm)^5(ak)^2, C(7, 2)(ak)b(bl)^4, and C(7, 2)a(bl)^2(ak)c^4.

Simplifying these expressions, we get:

C(7, 2)c^5a^2k^2m^5, C(7, 2)a^3kb^5l^4, and C(7, 2)ab^2kl^2c^4.

Notice that all these terms contain a^2, b^2, and c^2, as well as additional factors of a, b, and c.

Therefore, these terms are divisible by abc.

Since all terms in the expansion of (a + b + c)^7 that contain a, b, and c are divisible by abc, we can conclude that (a + b + c)^7 is divisible by abc.

Hence, abc divides (a + b + c)^7, as required.