Solution:

Given that the sum of the roots of the quadratic equation is 3 and the sum of the cubes of the roots is 7.

Let the roots of the quadratic equation be α and β.

According to the problem,

α + β = 3

α3 + β3 = 7

We know that,

α3 + β3 = (α + β) (α2 - αβ + β2)

Substituting the values from the given equations,

7 = 3 (α2 - αβ + β2)

α2 - αβ + β2 = 7/3

Now, we also know that the sum of the roots of the quadratic equation is given by

α + β = -b/a

where a and b are the coefficients of x2 and x in the quadratic equation.

Substituting the values from the given equation,

α + β = -(-b/a) = b/a

b/a = 3

b = 3a

We need to find the quadratic equation. Let the quadratic equation be

ax2 + bx + c = 0

We know that the sum of the roots of the quadratic equation is given by

α + β = -b/a

Substituting the values of b and a, we get

α + β = -3a/a = -3

α + β + 3 = 0

We also know that the sum of the product of the roots taken two at a time is given by

αβ = c/a

Substituting the values of a and b, we get

αβ = c/a = 1/3

αβ - c/a = 0

Multiplying both sides by 3a, we get

3αβ - 3c = 0

αβ = c/3

Substituting the values of αβ and α + β in the equation α2 - αβ + β2 = 7/3, we get

α2 - (c/3) + β2 = 7/3

Multiplying both sides by 3, we get

3α2 - c + 3β2 = 7

We also know that

α2 + β2 = (α + β)2 - 2αβ

Substituting the values of α + β and αβ, we get

α2 + β2 = 32 - 2(c/3)

3α2 + 3β2 = 9 - 2c

Adding this equation to the equation 3α2 - c + 3β2 = 7, we get

6α2 + 6β2 - c = 16

Substituting the value of c in this equation, we get

6α2 + 6β2 - (αβ) = 16/3

Multiplying both sides by 3, we get

18α2 + 18β2 - αβ = 16

Substituting the values of αβ and α + β in the equation α2 - αβ + β2 = 7/3, we get

6α2 + 6β2 - (α2 + β2 - (7/3)) = 16

5α2 + 5β2 = 31/3

Multiplying both sides by 3, we get