How to use Chinese Remainder Theorem A cubic polynomial $f(x)=ax^3+bx^2+cx+d$ gives remainders $-3x+3$ and $4x-1$ when divided by $x^2+x+1$ and $x^2+2x-4$. Find the value of $a,b,c,d$.
I know it’s easy but i wanna use Chinese Remainder Theorem(and Euclidean Algorithm) to solve it. 
A hint or a detailed answer would be much appreciated 

Step 1: Using extended Euclidean algorithm, find $g(x)$ and $h(x)$ such that $$g(x)(x^2+2x-4)+h(x)(x^2+x+1)=1$$.
\begin{align}
x^2+2x-4=x^2+x+1+(x-5)\implies& \begin{cases} q = 1\\
r=x-5\\
s=1\\
t=-1\end{cases}\\
x^2+x+1=(x+6)(x-5)+31\implies &\begin{cases} q = x+6\\
r=31\\
s=-(x+6)\\
t=1-(x+6)(-1)=x+7\end{cases}\\
 x-5=\left(\frac{x-5}{31}\right)31\implies& \begin{cases} q = \frac{x-5}{31}\\
r=0\\
s= 1+\frac{x-5}{31}(x+6)=\frac{x^2+x+1}{31}\\
t=-1-\frac{x-5}{31}(x+7)=\frac{x^2+2x-4}{31}\end{cases}
\end{align}
Therefore, $g(x)=-\frac{x+6}{31}$ and $h(x)=\frac{x+7}{31}$.
Step 2: Using Chinese remainder theorem, 
\begin{align}
f(x)&= (−3+3) + [(4−1)-(−3+3)]h(x)(x^2+x+1)\\&=(−3+3) +(7^3+3x^2+3x−4)\frac{(x+7)}{31}\\
&=\frac{7}{31}x^4+\frac{52}{31}x^3+\frac{24}{31}x^2-\frac{76}{31}x+\frac{65}{31}
\end{align}
