We know that since f(x) is divisible by x3, it can be written in the form f(x)=ax5+bx4+cx3+dx2+ex+f.

We are also given that f(x)+3 is divisible by (x+1)3, which can be expanded as (x+1)(x+1)(x+1)=x3+3x2+3x+1.

This tells us that there exist polynomials q(x) and r(x) such that:

f(x)+3=(x3+3x2+3x+1)q(x)+r(x)

Since (x3+3x2+3x+1) has a degree of 3, and we are given that f(x)+3 is divisible by it, then the remainder r(x) must be a polynomial of degree at most 2

(because any higher degree term in r(x) would be canceled out by the division). This means r(x)=mx2+nx+p for some constants m, n, and p.

Substituting the form of f(x) and expanding the equation, we get:

ax5+bx4+cx3+dx2+ex+f+3=(x3+3x2+3x+1)q(x)+mx2+nx+p

Matching coefficients of like terms on both sides, we get the following system of equations:

a=0 (since the coefficient of x5 on the right side is 0)

b=m+3q(1)

c=n+3q(0)+1 (since the constant term on the right side is 1)

d=p+3q(−1)

e=3q(−2)

f=p+3q(−3)

Since we are looking for the smallest possible value of f, we want to minimize the constant term p. From equation (6), we see that f is minimized when p is

minimized.

Setting p=0 and solving the remaining equations, we get:

a=0

b=m+3

c=n+1

d=−3

e=−6

f=−9

Therefore, the polynomial f(x) that satisfies the given conditions is:

f(x)=4x5+(m+3)x4+(n+1)x3−3x2−6x−9

The final step is to find the values of m and n that minimize f(x). Since we can choose any values for m and n as long as p=0, we can simply set them to 0.

This results in the polynomial:

f(x) = 4x^5 + 3x^4 + x^3