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 
 A: Not sure about the usage of the formulas, one obvious way is to write
$$ax^3+cx^2+cx+d$$
$$=(ax+e)(x^2+x+1)+(-3x+3)$$
$$=(ax+f)(x^2+2x-4)+(4x-1)$$
Compare the coefficients of the different exponents of $x$ to find the unknowns
As for example, comparing the constants,
$$d=e+3=-4f-1$$
A: EDIT: I just noticed my answer below doesn't meet OP's expectations, as I'm not explicitely using the Chinese theorem. I haven't read the question carefully enough...  

Reduction of $f(x)$ modulo $x^2+x+1$ can be computed by successively replacing all instances of $x^2$ in $f$ by $-x-1$, until we obtain an expression of degree $<2$. The process reduces $f(x)$ to $a(-x-1)x +b(-x-1) +cx +d=-ax^2+(c-a-b)x+d-b$, which in turns gives $ -a(-x-1)+(c-a-b)x+d-b = (c-b)x+a+d-b$.  
The same method with $x^2+2x-4$ tells us that $f(x)$ reduces to $(c-2b+8a)x-8a+d$. 
These two reductions must be respectively $-3x+3$ and $4x-1$. By very definition of equality between polynomials (that is, we can identify the coefficients), it reduces to the following system with $4$ equations and $4$ unknowns, which I let your solve by yourself:
$$\left\{
\begin{array}{}
-3&=c-b\\
3&=a+d-b\\
4&=c-2b+8a\\
-1&=-8a+d\\
\end{array}
\right.$$
A: $$  \left(   x^{2}  +  x  + 1 \right)  \left(   \frac{  -  x  - 7 }{ 31 }  \right)  -  \left(   x^{2}  + 2 x  - 4 \right)  \left(   \frac{  -  x  - 6 }{ 31 }  \right)  =  \left( -1  \right)  $$
is all you need. Cleaning up,
$$  (x+7) \left(   x^{2}  +  x  + 1 \right) - (x+6)  \left(   x^{2}  + 2 x  - 4 \right) = 31 $$
There is no guarantee of integer coefficients, even when both polynomials have integer coefficients. The units in $\mathbb Q[x]$ are nonzero rational constants.
A: 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}
