The original problem actually told that these six numbers are different. If $a,b,c,d,e,f$ are the six roots We have $abcdef=acf$ So $bde=1$ No three different integers would have a product of 1.

Mathlover_SH wrote: We have $abcdef=acf$ So $bde=1$ And what about $acf=0$ ?

The answer is no. Assume there are such $a, b, c, d, e, f$. Then \[(x + a)(x^2 + bx + c)(x^3 + dx^2 + ex + f) = k(x - a)(x - b)(x - c)(x - d)(x - e)(x - f)\]for some $k \ne 0$. Comparing coefficients of $x^6$, we see that $k = 1$. Then plugging in $x = 0$, we have $acf = abcdef$, implying $acf = 0$ or $bde = 1$. Since $b, d, e$ must be distinct, $bde = 1$ is impossible, so $acf = 0$. Case 1: $a = 0$. Then $(x^2 + bx + c)(x^3 + dx^2 + ex + f) = (x - b)(x - c)(x - d)(x - e)(x - f)$. Plugging in $x = 0$, we have $cf = -bcdef$, implying $bde = -1$, since $cf \ne 0$. But $b, d, e$ are distinct, so $bde = -1$ is impossible. So no solutions in this case. Case 2: $c = 0$. Then $(x + a)(x + b)(x^3 + dx^2 + ex + f) = (x - a)(x - b)(x - d)(x - e)(x - f)$. Plugging in $x = 0$, we have $abf = -abdef$, implying $de = -1$, since $abf \ne 0$. Then $(d, e) = (1, -1)$ or $(-1, 1)$. So $a, b \notin \{-1, 0, 1\}$. Then $-a, a, -b, b, -1, 1, 0$ are all distinct roots, contradiction. So no solutions in this case. Case 3: $f = 0$. Then $(x + a)(x^2 + bx + c)(x^2 + dx + e) = (x - a)(x - b)(x - c)(x - d)(x - e)$. Plugging in $x = 0$, we have $ace = -abcde$, implying $bd = -1$, since $ace \ne 0$. WLOG let $b = 1$ and $d = -1$ (since if $(a, 1, c, -1, e, 0)$ is a solution, then $(a, -1, e, 1, c, 0)$ is as well). Then $(x + a)(x^2 + x + c)(x^2 - x + e) = (x - a)(x - c)(x - e)(x + 1)(x - 1)$. Plugging in $x = 1$, we have $e(a + 1)(c + 2) = 0$, since $a \ne -1$ and $e \ne 0$, we must have $c = -2$. Similarly, by plugging $x = -1$, we get $e = -2$, contradiction. So no solutions in this case. In all cases, no $a, b, c, d, e, f$ satisfy the conditions. $\Box$