Q.118. Simultaneous equations

Question 118.
Solve
a + b + c = 6
a² + b² + c² = 14
a³ + b³ + c³ = 36

Answer 118.
a + b + c = 6   ...   (1)
a² + b² + c² = 14   ...   (2)
a³ + b³ + c³ = 36   ...   (3)

Squarring eqn. (1) and subtracting eqn. (2) from it,
2 (ab + bc +ca) = 22
=> ab + bc+ ca = 11

a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)
=> 36 - 3abc = 6 (14 - 11)
=> abc = 6

Let a, b, c be the roots of a cubic polynomial
=> (x - a) (x - b) (x - c) = 0
=> x^3 - (a + b + c) x^2 + (ab + bc + ca) x - abc = 0
=> x^3 - 6x^2 + 11x - 6 = 0
 x = 1 satisfies the equation
=> (x - 1) is its solution.

Rearranging terms of polynomial keeping in mind that (x - 1) is its solution,
=> x^3 - x^2 - 5x^2 + 5x + 6x - 6 = 0
=> x^2 (x - 1) - 5x (x - 1) + 6 (x - 1) = 0
=> (x - 1) (x^2 - 5x+ 6) = 0
=> (x - 1) (x - 2) (x - 3) = 0
=> 1, 2 and 3 are the roots.

As changing a, b and c in cyclic order in the above equations does not change the equations,
any one of a, b and c can be taken as 1, any other as 2 and the third as 3.
Thus, there are six sets of solutions:
(1, 2, 3), (2, 3, 1), (3, 1, 2), (1, 3, 2), (3, 2, 1) and ( 2, 1, 3).

The general solution of the equations
a + b + c = p   ...   (1),   a² + b² + c² = q   ...   (2)   and   a³ + b³ + c³ = r   ...   (3) is the roots of the polynomial,
x^3 - px^2 + (1/2)(p^2 - q)x - (1/6)(2r - 3pq + p^3) = 0.
If one of the roots is a simple small integer, the equation can be solved manually, otherwise use Wolfram Alpha.

Try to solve the following set of equations using the above general formula.
a + b + c = 3 ... (1),   a² + b² + c² = 3.5... (2)   a³ + b³ + c³ = 4.5... (3)

LINK to YA!