<s>[INST] If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is: 
If $a$, $b$ are the roots of $x^2-2x+3$.Then the equation whose roots are $a^3-3a^2+5a-2$ and $b^3-b^2+b+5$ is:   

I have not been able to find a better method than to calculate $a$ and $b$ then substitute them into the roots for the new polynomial. 
I believe this question can't be transformed in a similar manner as mentioned in this question as the new roots are asymmetrical.  
Does a better method than the lackluster substitution, exist?  
The answer is:

 $x^2-3x+2$

 [/INST] Since $x^2=2x-3$, we get that $x^3=2x^2-3x=x-6$. Then,
$$
\begin{align}
a^3-3a^2+5a-2
&=(a-6)-3(2a-3)+5a-2\\
&=1
\end{align}
$$
and
$$
\begin{align}
b^3-b^2+b+5
&=(b-6)-(2b-3)+b+5\\
&=2
\end{align}
$$
It is easy to find an equation which has roots of $1$ and $2$.
</s>