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 need to show that (a + b + c)^7 is divisible by abc. We will do this by considering the prime factorization of a, b, and c, and using the properties of divisibility and the binomial theorem.

Step 1: Prime Factorization
Let's express a, b, and c in terms of their prime factorizations:
a = p1^x1 * p2^x2 * ... * pn^xn
b = q1^y1 * q2^y2 * ... * qm^ym
c = r1^z1 * r2^z2 * ... * rk^zk

Here, p1, p2, ..., pn are the distinct prime factors of a, q1, q2, ..., qm are the distinct prime factors of b, and r1, r2, ..., rk are the distinct prime factors of c. The exponents x1, x2, ..., xn, y1, y2, ..., ym, z1, z2, ..., zk represent the respective powers of these prime factors in the prime factorization of a, b, and c.

Step 2: Divisibility Relations
From the given conditions:
- a divides b^2 ⇒ All prime factors of a are present in b^2, and their respective exponents in b^2 are at least double those in a.
- b divides c^2 ⇒ All prime factors of b are present in c^2, and their respective exponents in c^2 are at least double those in b.
- c divides a^2 ⇒ All prime factors of c are present in a^2, and their respective exponents in a^2 are at least double those in c.

Step 3: Exponent Inequalities
Based on the divisibility relations, we can write the following inequalities for the exponents of the prime factors:
2x1 ≤ 2y1 ≤ 2z1 ≤ 2x2 ≤ 2y2 ≤ 2z2 ≤ ... ≤ 2xn ≤ 2yn ≤ 2zn
To simplify, let's assume that the exponents are in increasing order:
x1 ≤ y1 ≤ z1 ≤ x2 ≤ y2 ≤ z2 ≤ ... ≤ xn ≤ yn ≤ zn

Step 4: (a + b + c)^7
Now, let's expand (a + b + c)^7 using the binomial theorem:
(a + b + c)^7 = Σ (nCr * a^(7-r) * b^r * c^(7-r))

In each term of the sum, a^(7-r) represents the contribution of the prime factors present in a to the power of 7-r, b^r represents the contribution of the prime factors present in b to the power of r, and c^(7-r) represents the contribution of the prime factors present in c to the power of 7-r.

Step 5: Divisibility of (a + b + c)^7 by abc
To prove that abc divides (a + b + c)^7, we need to show that each term of the expansion, Σ (nCr * a^(7-r) * b^r * c^(7-r)), is divisible by abc.

From Step 2, we know that prime factors in a are present in b^2, prime factors in b are present in c^2, and prime factors in c are present in a^2.

Considering an individual term of the expansion, Σ (nCr * a^(7-r) * b^r * c^(7-r)), we can conclude:

1. The prime factors in a^(7-r) are all present in (b^2)^(7-r) = b^(14-2r).
Thus, a^(7-r) is divisible by b^(14-2r).

2. The prime factors in b^r are all present in (c^2)^r = c^(2r).
Thus, b^r is divisible by c^(2r).

3. The prime factors in c^(7-r) are all present in (a^2)^(7-r) = a^(14-2r).
Thus, c^(7-r) is divisible by a^(14-2r).

From the above observations, we see that each term of the expansion is divisible by abc, as:
(a^(7-r) is divisible by b^(14-2r)) * (b^r is divisible by c^(2r)) * (c^(7-r) is divisible by a^(14-2r)) = abc

Therefore, (a + b + c)^7 is divisible by abc.

Hence, we have proved that abc divides (a + b + c)^7.