Find the values of a and b in this cubic

Two of the roots are $-2$ and $-3$ . Call the third root $r$ .

Using Vieta's formula , $1=r_1r_2+r_1r_3+r_2r_3=(-2)(-3)+(-2)r+(-3)r$ so $r=1$ . Thus $a=-(r_1+r_2+r_3)=-(-2+-3+1)=4$ and $b=-(r_1r_2r_3)=-((-2)(-3)(1))=6$ , yielding $a+b=4-6=-2$

Let (x+2)(x+3)(x+c) = x^3+ax^2+x+b

Then LHS is (x^2+5x+6)(x+c)

x^3+(5+c)x^2+(6+5c)x +6c

Comparing:

6 + 5c=1

5c = −5

c = −1

a = 2 + 3 + c = 5−1=4

b =6c = 6(−1) = -6.

So a = 4, b = −6

a + b = -2