\(f\left(x\right)=x^3-2ax+a^2\)
\(\Rightarrow f\left(1\right)=1-2a+a^2\)
\(g\left(x\right)=x^3+\left(3a+1\right)x+a^2\)
\(\Rightarrow g\left(3\right)=27+\left(3a+1\right)3+a^2\)
Mà \(f\left(1\right)=g\left(3\right)\)
\(\Rightarrow1-2a+a^2=27+\left(3a+1\right)3+a^2\)
\(\Rightarrow1-2a=27+9a+3\)
\(\Rightarrow1-2a=30+9a\)
\(\Rightarrow-29=11a\)
\(\Rightarrow a=\dfrac{-29}{11}\)
Vậy \(a=\dfrac{-29}{11}\) thì \(f\left(1\right)=g\left(3\right)\)