simplifying $\frac{15^4+105^2-225\cdot210}{130^2-50^2}$ $x=\frac{15^4+105^2-225\cdot210}{130^2-50^2}$
Since $225 = 15^2$ and $210 = 2\cdot105$ the numerator is $(15^2-105)^2$ which is $120^2 = 14400$.
The denominator is $(130-50)(130+50)$ which is $14400$. The answer is $1$. The solutions for the exam have proven to be wrong from time to time and in this case they state that $x=10$. This is wrong answer, or?

$\dfrac{15^4+105^2-225\cdot210}{130^2-50^2}=\dfrac{(15^2-105)^2}{(130+50)(130-50)}=\dfrac{120^2}{180\times80}=\dfrac{12^2}{18\times8}=1$
