Cho g_{( x )} = 0

\(\Leftrightarrow\)( x - 2 )( x - 3 ) = 0

\(\Leftrightarrow\)x = 2 hoặc x = 3

f_{( 2 )} = 2 . 2³ - 3 . a . 2² + 2 . 2 + b = 20 - 12a + b ( 1 )

f_{( 3 )} = 2 . 3³ - 3 . a . 3² + 2 . 3 + b = 48 - 27a + b ( 2 )

Lấy ( 1 ) và ( 2 ) ta có :

   - 28 + 15a = 0

\(\Rightarrow\)15a = 28 

\(\Rightarrow\)a = 28 / 15

\(\Rightarrow\)b = 12 / 5

f(3)=g(1)

nên \(1+3\left(3a+1\right)+a^2=1-2a+a^2\)

\(\Leftrightarrow1+9a+3=1-2a\)

=>11a=-3

hay a=-3/11

\(2x^3-3ax^2+2x+b⋮\left(x-1\right)\left(x+2\right)\)

\(\Leftrightarrow2x^3-3ax^2+2x+b⋮x^2+x-2\)

\(\Leftrightarrow2x^3+2x^2-4x+2x^2+2x-4+\left(-4-3a\right)x^2+\left(b+4\right)⋮x^2+x-2\)

=>-3a-4=0 và b+4=0

=>a=-4/3 và b=-4

\(f\left(x\right)=x^3-2ax+a^2\)

\(\Rightarrow f\left(1\right)=1-2a+a^2\)

\(g\left(x\right)=x^3+\left(3a+1\right)x+a^2\)

\(\Rightarrow g\left(3\right)=27+\left(3a+1\right)3+a^2\)

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 \(f\left(1\right)=g\left(3\right)\)