Find and simplify $a^3+c^3-b^3c\pmod{19}$ when $a \equiv 2\pmod {19},b \equiv 4\pmod {19}, c \equiv 5\pmod {19}$

Question

Suppose $a \equiv 2\pmod {19},b \equiv 4\pmod {19}, c \equiv 5\pmod {19}$. Find and simplify $a^3+c^3-b^3c\pmod{19}$

My attempt

$a^3+c^3-b^3c\pmod{19}=2^3+5^3-(4^3*5)\pmod{19}=8+125-320\pmod{19}=8+125-16=117$

Am I on the right track?

Answer 1

Your idea is correct, but the notation is a bit redundant. In one line, you can have $$\begin{align}a^3+c^3-b^3c&\equiv2^3+5^3-4^3(5)\equiv8+125-320\equiv-187\equiv3\pmod{19}\end{align}$$