The real numbers $a_1,a_2,a_3$ and $b{}$ are given. The equation \[(x-a_1)(x-a_2)(x-a_3)=b\]has three distinct real roots, $c_1,c_2,c_3.$ Determine the roots of the equation \[(x+c_1)(x+c_2)(x+c_3)=b.\][i]Proposed by A. Antropov and K. Sukhov[/i]
1. We start with the given equation:
   \[
   (x - a_1)(x - a_2)(x - a_3) = b
   \]
   This equation has three distinct real roots \(c_1, c_2, c_3\). Therefore, we can write:
   \[
   (c_1 - a_1)(c_1 - a_2)(c_1 - a_3) = b
   \]
   \[
   (c_2 - a_1)(c_2 - a_2)(c_2 - a_3) = b
   \]
   \[
   (c_3 - a_1)(c_3 - a_2)(c_3 - a_3) = b
   \]

2. We need to determine the roots of the equation:
   \[
   (x + c_1)(x + c_2)(x + c_3) = b
   \]

3. Consider the polynomial \(P(x) = (x - a_1)(x - a_2)(x - a_3) - b\). The roots of \(P(x) = 0\) are \(c_1, c_2, c_3\).

4. Now, consider the polynomial \(Q(x) = (x + c_1)(x + c_2)(x + c_3) - b\). We need to find the roots of \(Q(x) = 0\).

5. Notice that if we substitute \(x = -a_i\) (for \(i = 1, 2, 3\)) into \(Q(x)\), we get:
   \[
   Q(-a_i) = (-a_i + c_1)(-a_i + c_2)(-a_i + c_3) - b
   \]

6. Since \(c_1, c_2, c_3\) are the roots of the polynomial \(P(x)\), we have:
   \[
   (c_1 - a_1)(c_1 - a_2)(c_1 - a_3) = b
   \]
   \[
   (c_2 - a_1)(c_2 - a_2)(c_2 - a_3) = b
   \]
   \[
   (c_3 - a_1)(c_3 - a_2)(c_3 - a_3) = b
   \]

7. Therefore, substituting \(x = -a_i\) into \(Q(x)\) gives:
   \[
   Q(-a_i) = (-a_i + c_1)(-a_i + c_2)(-a_i + c_3) - b = (c_1 - a_i)(c_2 - a_i)(c_3 - a_i) - b = b - b = 0
   \]

8. Hence, \(-a_1, -a_2, -a_3\) are the roots of the equation \((x + c_1)(x + c_2)(x + c_3) = b\).

The final answer is \(\boxed{-a_1, -a_2, -a_3}\).