JOMO 6, Short 7

$(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$

$\Rightarrow ab+bc+ca = \frac{(a+b+c)^2 - a^2+b^2+c^2}{2} = \frac{6^2-66}{2} = -15$

$$

$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\Rightarrow 666-3abc = 6 \times (66+15)$

$\Rightarrow abc = 60$

$$

$(ab+bc+ca)^2 =a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2$

$\Rightarrow (ab+bc+ca)^2 = a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)$

$\Rightarrow (-15)^2 = a^2b^2+b^2c^2+c^2a^2 + 2\times 60 \times 6$

$\Rightarrow a^2b^2+b^2c^2+c^2a^2 = -495$

$$

$(a^2+b^2+c^2)^2 = a^4+b^4+c^4+2(a^2b^2+b^2c^2+a^2c^2)$

$\Rightarrow 66^2 = a^4+b^4+c^4 + 2\times (-495)$

$\Rightarrow a^4+b^4+c^4 = 5346$

Last 3 digits are $\boxed{346}$

Consider the cubic $(x-a)(x-b)(x-c)$ , which can be expressed as $x^3+p_1x^2+p_2x+p_3$ . By Newton's Sums, we have $6+p_1 = 0 \rightarrow p_1=-6$ . This leads us to obtain $66-36+2a_2 = 0 \rightarrow p_2=-15$ , and $666-396-90+3a_3=0 \rightarrow p_3=-60$ . Thus, $a$ , $b$ , and $c$ are the roots of the polynomial $x^3-6x^2-15x-60=0$ . We now apply Newton's Sums once more to obtain $a^4+b^4+c^4-3996-990-360=0 \rightarrow a^4+b^4+c^4=5346$ and the last $3$ digits are $\boxed{346}$ .