Let f(x) = (x³ + ax² + bx + 6)

By using remainder theorem,

If we divide f(x) by (x – 3) then it will leave a remainder as f(3)

So,

f(3) = 3² + a × 3² + b × 3 + 6 = 3

27 + 9a + 3b + 6 = 3

9a + 3b + 33 = 3

9a + 3b = 3 – 33

9a + 3b = -30

3a + b = -10 (i)

It is also given that,

(x – 2) is a factor of f(x)

Therefore,

By using factor theorem, we get

(x – a) is the factor of f(x) if f(a) = 0and also f(2) = 0

Now,

f(2) = 2³ + a × 2² + b × 2 + 6 = 0

8 + 4a + 2b + 6 = 0

4a + 2b = -14

2a + b = -7 (ii)

Now by subtracting (ii) from (i), we get

a = -3

Putting the value of a in (i), we get

3(-3) + b = -10

-9 + b = -10

b = -10 + 9

b = -1

Therefore,

b = -1 and a = -3