[INST] Given that a+b=3 and a^3+b^3=81, find ab./n [/INST]  ###Response: We are given a system of equations:

a + b = 3     -------------- (Equation 1)

a³ + b³ = 81  -------------- (Equation 2)

To find the value of ab, we can use the sum of cubes factorization, which is a³ + b³ = (a + b)(a² - ab + b²). Using this, we can rewrite Equation 2 as:

(a + b)(a² - ab + b²) = 81

Substituting Equation 1, we get:

3(a² - ab + b²) = 81

Simplifying further, we get:

a² - ab + b² = 27   -------------- (Equation 3)

We can now square Equation 1 to get:

(a + b)² = a² + 2ab + b² = 9

Using Equation 3, we can rewrite this as:

a² + 2ab + b² = (a² - ab + b²) + 2ab + ab = 27 + 3ab = 9

Simplifying, we get:

3ab = -18

Therefore, ab = -6.