If the sum of the roots of a quadratic equation is 3 and the sum of their cubes is 63 find the equation.

Assume the roots of the equation we have to find are r1
and r2.

So (x-r1)(x-r2)=0

or
x^2-(r1+r2)x+r1*r2=0

Let's write this as
x^2-bx+c=0

Here b=r1+r2 and
c=r1*r2.

Now it is given that the sum of the roots is 3,
therefore b=3.

And as the sum of their cubes is 63,
r1^3+r2^3=63.

Now
(r1+r2)^3=r1^3+r2^3+3*r1*r2*(r1+r2).

Substituting the
values we have, 3^3=63+9*r1*r2.

Therefore r1*r2=
(27-63)/9=-4. Or c in the quadratic equation is equal to
-4.

The required quadratic equation is
x^2-3x-4=0

To check: this quadratic equation
has roots 4 and -1. -1+4=3 and (-1)^3+4^3=63.