Given,
f(x)=3x3+2bx+2ax2 and g(x)=3x3+ax+bx2
Now differentiating the above function we get,
f′(x)=x2+2b+ax
g′(x)=x2+a+2bx
Now given, f′(x)=0andg′(x)=0 have a common extreme point,
So, x2+2b+ax=x2+a+2bx
⇒(2b−a)−x(2b−a)=0
∴x=1 is the common extreme point
Put x=1 in f′(x)=0 or g′(x)=0 we get,
1+2b+a=0
⇒a+2b+7=6