Find...

$\begin{aligned} a^3+b^3+c^3-3abc & = & (a+ b+c)(a^2+b^2+c^2-ab-bc-ac) \\ & = & (a+b+c) \left ( (a+b+c)^2-2(ab+ab+bc) - (ab+ac+bc) \right ) \\ & = & (a+b+c) \left ( (a+b+c)^2-3(ab+ab+bc) \right ) \\ & = & (11) \left ( (11)^2-3(25) \right ) \\ & = & \boxed{506} \\ \end{aligned}$

$\displaystyle \begin{aligned} a^3+b^3+c^3-3abc &= (a+b+c) \left((a+b+c)^2 - 3(ab+bc+ca) \right) \\ &= 11 \times (11^2 - 3 \times 25) \\ &= 506 \end{aligned}$

(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac) = ----> 11^2 = a^2 + b^2 + c^2 + 2(50)----> a^2 + b^2 + c^2 = 121 - 50 = 71 ---> Next, (a^2 + b^2 + c^2)(a+b+c) = a^3 + b^3 + c^3 + a^2.b + a^2.c + b^2.a + b^2.c + c^2.a + c^2.b ----> 71.11 = a^3 + b^3 + c^3 + ab(a+b) + bc(b+c) + ac(a+c) ---> 781 = a^3 + b^3 + c^3 + ab(11-c) + bc(11-a) + ac(11-b) ---> 781 = a^3 + b^3 + c^3 - 3.abc +11(ab + bc + ac) ---> 781 = a^3 + b^3 + c^3 - 3.abc +11.25 ----> 781-275 = a^3 + b^3 + c^3 - 3.abc ----> a^3 + b^3 + c^3 - 3.abc = 781 - 275 = 506. ANSWER : 506