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:

 $x^2-3x+2$


Hint $\ x^3-3x^2+5x-2\,\bmod\, \color{#c00}{x^2-2x+3}\, =\, \color{#0a0}1\ $ (and $= \color{#90f}2$ for the other). So we seek a polynomial with roots $\color{#0a0}1$ and $\color{#90f}2,\,$ e.g. $\ (x-\color{#0a0}1)(x-\color{#90f}2)$
Remark $ $ The remainder is quickly computable by long division (ignoring the unneeded quotient)
$$\begin{align}
&\ \ \ 1\ {-}3\ \ \ \, 5\,\ {-}2\\
&\color{#c00}{{-}1\,\ \ \  2\ {-}3}\\
& \ \ \ \ \ \ {-}1\ \ \ \ 2\ \ {-}2\\
&\color{#c00}{\ \ \ \ \ \ \ \ \    1\ {-}2\ \ \ \ \ 3}\\
&\qquad\qquad\quad\ \  \color{#0a0}1
\end{align}\qquad\qquad\quad$$
