Let, f(x) = x^9 - 6*x^7 + 9*x^5 - 4*x^3

Then, factorizing f(x), we have:

f(x) = x^3*(x+1)^2*(x-1)^2*(x+2)*(x-2)

If x = 3p, then f(x) is divisible by (3p)^3 = 27*p^3

If x= 3p+1, then f(x) is divisible by (3p)^2*(3p+3) = 27*(p^2)*(p+1)

If x = 3p+2, then f(x) is divisible by (3p+3)^2*(3p) = 27*p*(p+1)^2

Thus, *f(x) is always divisible by 27*......(i)

Again, f(x) = x^2(x^2 -1)*g(x), where:

g(x) = (x+2)(x+1)x(x-1)(x-2).

Since f(x) is divisible by the product of five consecutive integrs, it follows that f(x) must *always be divisible by 5*.......(ii)

If x is even, then either x = 4p, or (4p+2)

For x = 4p, f(x) is divisible by (4p)^3*(4), so that f(x) is divisible by 256

For x = 4p+2, f(x) is divisible by 128*(2p+1)^3*p(p+1), so that f(x) is divisible by 128*2 = 256.

Accordingly, f(x) is divisible by 256, whenever f(x) is even.

If x is odd, then x^2-1 is divisible by 8. Let (x^2-1)/8 = M (say). Then, writing:

f(x)= x^3*(x^2-1)^2*(x^2-4), we observe that:

f(x) is divisible by 64*M^2, so that f(x) is divisible by 64, whenever x is odd.

Thus, *f(x) is always divisible by 64*......(iii)

Combining (i), (ii) and (iii), it follows that in general *f(x) is an exact multiple of * LCM(27,5,64) =* 8,640*.

*Edited on ***September 8, 2008, 6:10 am**