**Solution :** Comparing the given polynomial with ax^{3} + bx^{2} + cx + d, we get

a = 3, b = – 5, c = –11, d = – 3. Further

p(3) = 3 × 3^{3} – (5 × 3^{2}) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,

p(–1) = 3 × (–1)^{3} – 5 × (–1)^{2} – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,

p(-1/3) = 3* (-1/3)^{3} - 5*(-1/3)^{2} - 11*(-1/3) - 3,

= -1/9-5/9+11/3-3 = 0

Therefore, 3, -1 and -1/3 are the zeroes of 3x^{3} - 5x^{2} -11x -3

So, we take α = 3, β = -1, γ = -1/3,

Now,

α + β + γ = 5/3 = -(-5)/3 = -b/a,

αβ + βγ + γα = -11/3 = c/a,

αβγ = 1 = -(-3)/3 = -d/a