Posted by Meenakshi on .
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 .
maths --plse help me.. -
(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
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