f'' = 56*x^6 - 42*x^5 + 2

Its minimum value occurs when its own derivative is zero which is when 56*6*x^5 - 42*5*x^4 = 0 which happens when x is zero and when x = 42 * 5 / (56 * 6) = 5/8. When x is zero f'' is 2 and so is positive, and when x = 5/8 

f'' = 2 + 7*8*(5/8)^6 - 7*6*(5/8)^5

    = 2 + (5/8)^5 * (7 * 5 - 7 * 6) = 2 - 7*(5/8)^5  which is ~ 1.33

Since the minimum of f'' is positive, it's always positive.

Because the second derivative is always positive, there can't be more than one place where the first derivative is zero (or else the first derivative would have a maximum or minimum between those multiple zeros). Where is that place?

Well, if x is greater than 1 then x^8 increases faster than x^7 and x^2 increases faster than x^1 so f(x) is increasing. Similarly, if x is < 0 then x^8 and x^2 are increasing and x^7 and x are decreasing but because they're being subtracted they also result in an increase. So any minimum must lie in the range [0,1]

In that range, the smallest x^n - x^(n-1) can be is when x = (n-1)/n, and is - (n-1)^(n-1)/n^n.

So the smallest value the first two terms of f can have is -7^7/8^8 which is ~ -0.05 and the smallest the next two terms can have is -1^1/2^2 = -0.25. And that means the smallest the total can be is around -.3 (these minima aren't at the same value of x so the actual minimum is larger, but that doesn't matter.)

Since the overall minimum of f is therefore  > ~1.7, there are no zeros.