An algebra problem by Fidel Simanjuntak

$\begin{aligned} ( x-y )^2 & = x^2 + y^2 - 2xy \\ x^2 + y^2 & = (x-y)^2 + 2xy \\ x^2 + y^2 & = 100 + 20 = 120 \\ \text{Now,} \\ (x^2 + y^2)^2 & = x^4 + y^4 + 2x^2y^2 \\ x^4 + y^4 & = (x^2 + y^2)^2 - 2x^2y^2 \\ & = 14400 - 200 = \boxed{14200} \end{aligned}$

$\begin{aligned} (x-y)^4 & = 10^4 \\ x^4-4x^3y+6x^2y^2-4xy^3+y^4 & = 10000 \\ x^4+y^4-4xy\left(x^2+y^2\right)+6x^2y^2 & = 10000 \\ x^4+y^4-4xy\left((x-y)^2+2xy\right)+6(xy)^2 & = 10000 \\ x^4+y^4-4(10)\left(10^2+2(10)\right)+6(10)^2 & = 10000 \\ x^4+y^4-40\left(120\right)+600 & = 10000 \\ x^4+y^4-4200 & = 10000 \\ x^4+y^4 & = \boxed{14200} \end{aligned}$