Kill it with identities

Let a,b,c are three such numbers then, we are given $a+b+c = 6, a^2+b^2+c^2 = 6, a^3+b^3+c^3 = 6$ .
$(a+b+c)^2 = 36$ $\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca = 36$ $\Rightarrow 6+2(ab+bc+ca) = 36$
$\Rightarrow ab+bc+ca = 15$ .....(1)
Now, $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca).$

$\Rightarrow 6-3abc = 6\times (6-15). [from (1)]$
$\Rightarrow 3abc = 60 \Rightarrow abc = 20$

Letting $a, b, c$ be the numbers, one has

$36 = 6^2 = (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac = 6 + 2ab + 2bc + 2ac$ .

Thus $ab + bc + ac = 15$ . Let $P = abc$ .

The polynomial

$(x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab + bc + ac)x - abc = x^3 - 6x^2 + 15x - P$ .

Since this polynomial equals zero when $x=a,b,c$ , one has

$a^3 - 6a^2 + 15a - P =0$

$b^3 - 6b^2 + 15b - P = 0$

$c^3 - 6c^2 + 15c - P =0$ ,

and adding these equations gives

$(a^3 + b^3 + c^3) - 6(a^2 + b^2 + c^2) + 15(a+b+c) - 3P=0$

$6 - 36 + 90 - 3P=0$ ,

whence $P=20.$

It can simply be shown that: (a+b+c)^3+2 (a^3+b^3+c^3)-3 (a^2+b^2+c^2)(a+b+c)=6abc Since, (a+b+c), (a^3+b^3+c^3), (a^2+b^2+c^2) = 6 Then all terms on the left hand side within the brackets can be computed easily. Then after dividing through by 6 on both sides and computing the left hand side your left with 20=abc