Modular arithmatic $x^a \mod N= a\cdot b$ $x^3 \equiv 2 \pmod{15}$
How to solve it mod $3$ and mod $5$?
And how does the Chinese remainder theorem help?
I want a general method to follow in case the modulus is composite number. 
Thanks in advance!

$x^3\equiv 2\pmod{15}\Rightarrow \begin{cases}x^3\equiv 2\pmod{3}\\x^3\equiv 2\pmod{5}\end{cases}$
It is clear that if $x\equiv 0$ or $x\equiv 1\pmod{3}$ that $x^3\not\equiv 2\pmod{3}$ so we know that $x\equiv 2\pmod{3}$.  Checking confirms this since $2^3=8\equiv 2\pmod{3}$
For the second implication, it should be clear that $x\equiv 0,1,4\pmod{5}$ will result in $x^3\equiv 0,1,4\pmod{5}$ respectively.  We are left to check when $x\equiv 3$ and $2$.
$2^3=8\equiv 3\pmod{5}$
$3^3=27\equiv 2\pmod{5}$
Thus, we have the system of congruencies $\begin{cases}x\equiv 2\pmod{3}\\x\equiv 3\pmod{5}\end{cases}$
Applying chinese remainder theorem and solving the system yields $x\equiv 8\pmod{15}$
