Để $P(x)=Q(x)$
$⇒x^2-2ax+a^2=x^2+(3a-1)x+a^2$
$⇒x^2-2ax+a^2-x^2-(3a-1)x-a^2=0$
$⇒x^2-2ax+a^2-x^2-3ax+x-a^2=0$
$⇒(x^2-x^2)+(-2ax-3ax)+(a^2-a^2)+x=0$
$⇒-5ax+x=0$
$⇒x(-5a+1)=0$
\(⇒\left[ \begin{array}{l}x=0\\-5a+1=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=0\\a=\dfrac{1}{5}\end{array} \right.\)
Vậy với $x=0$ hoặc $a=\dfrac{1}{5}$ thì $P(x)=Q(x)$.