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 .
(a+b+c)^7 = (a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)
will be of the form
a^7 + c1*a^6*b + c2*a^6*c + c3*a^5*b^2 + c4*a^5*c^2 + c5*a^5*b*c + . . . .
where c1, c2, c3 are constants found when doing the actual expansion.
so all the terms in the expansion are of the form
ci*a^x*b^y*c^z
where x, y, z are integers from 0 to 7, and x+y+z = 7
So (ci*a^x*b^y*c^z)/abc = ci*a^(x-1)*b^(y-1)*c^(z-1)
So in the case when x, y, and z, are all 1 or greater, abc divides these terms in the expansion.
We are left with the terms when x, y, or z is 0
the terms involving
a^7, b^7, c^7, a^6*b, a^6*c, a*b^6, a*c^6, b^6*c, 6*c^6
a^7 / abc = a^6/bc; but c divides a^2; so we need to show that a^4/b is a real number; b divides c^2, so c^2 is a multiple of b; a^4/b if and only if a^4/c^2; c divides a^2; so c^2 divides a^4; a^7 divides abc
Use this type of logic for the rest of the terms b^7, . . .etc
To prove that abc divides (a + b + c)^7, we need to show that (a + b + c)^7 is divisible by abc. We will use the given conditions to prove this statement.
Since a divides b^2, we can write b^2 = ax for some positive integer x.
Similarly, b divides c^2, so we can write c^2 = by for some positive integer y.
Finally, c divides a^2, so we can write a^2 = cz for some positive integer z.
Now, let's consider the expression (a + b + c)^7. We can expand it using the binomial theorem:
(a + b + c)^7 = a^7 + 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 + b^7
+ 7a^6c + 42a^5bc + 105a^4b^2c + 140a^3b^3c + 105a^2b^4c + 42ab^5c + 7b^6c
+ 7a^6c + 7a^5bc + 21a^4b^2c + 35a^3b^3c + 35a^2b^4c + 21ab^5c + 7b^6c + c^7
Now, we will analyze each term of the expanded expression.
1. The term a^7 has factors of a * a * a * a * a * a * a. We have a factor of a from each of the terms a^6b, a^5bc, and a^6c. Since a, b, and c divide a^2, we know that they divide a * a, so they also divide a^7.
2. The term 7a^6b can be factored as 7 * a * a * a * a * a * a * b. We have a factor of a from each of the terms a^6b, a^5bc, and a^6c. Additionally, since a divides b^2, we can substitute b^2 = ax, giving us 7a^5(ax) = 7a^6x, which means that b also divides 7a^6b.
3. Following the same logic, we can see that each term of the expanded expression has factors of a, b, and c. Since a, b, and c divide a^2, b^2, and c^2 respectively, they will also divide a * a, b * b, and c * c. Therefore, each term is divisible by abc.
Combining all the factors, we conclude that all the terms of (a + b + c)^7 are divisible by abc. Hence, abc divides (a + b + c)^7.