The function F(x) can be expanded using the binomial theorem, which states that (a+b)^n = a^n + n(a^n-1)b + n(n-1)/2(a^n-2)b^2 + ... + b^n.

Applying this theorem to F(x)=3(x^5+7x^3+1)^4, we get:

F(x) = 3(x^5+7x^3+1)^4
= 3(x^5 + 7x^3 + 1)(x^5 + 7x^3 + 1)(x^5 + 7x^3 + 1)(x^5 + 7x^3 + 1)
= 3[(x^5)^4 + 4(x^5)^3(7x^3) + 6(x^5)^2(7x^3)^2 + 4(x^5)*(7x^3)^3 + (7x^3)^4 + ...

4(x^5)^3 + 12(x^5)^2(7x^3) + 12(x^5)(7x^3)^2 + 4(7x^3)^3 + ...6(x^5)^2 + 12(x^5)(7x^3) + 6(7x^3)^2 + ...4(x^5) + 4(7x^3) + ...1]

Simplifying the expression, we get:

F(x) = 3[x^20 + 28x^16 + 84x^14 + 112x^12 + 49x^12 + 12x^10 + 336x^8 + 84x^8 + 42x^6 + 28x^6 + 168x^4 + 12x^4 + 24x^2 + 4x^2 + 1]
= 3(x^20 + 28x^16 + 133x^14 + 124x^12 + 336x^8 + 112x^6 + 168x^4 + 36x^2 + 1)

Therefore, the expanded form of the function F(x) is 3x^20 + 84x^16 + 399x^14 + 372x^12 + 1008x^8 + 336x^6 + 504x^4 + 108x^2 + 3.