If we create the matrix of the system and after that we calculate it's determinant,we'll note that the determinant is Vandermonde type, that means that the solution is the result of multiplication of the 3 factors: (b-a)(c-a)(c-b) and this one is different from zero, if a different from b and different from c.

In this case , the system has just a single solution and it's determined. We could solve the system using the Cramer method, where:

x = det 1/det A

Y = det 2 / det A

z = det 3 / det A

det1, det2, det3 - determinants formed from the initial matrix of a system, where the column formed by the coefficients of the searched unknown is substituted by the column of the free terms.

det A - determinant of system matrix.

det1, det2, det3 - also Vandermonde determinants.

det1 = (b-1)(c-1)(c-b) (we've substituted the column formed by the coefficients 1, a and a^2 with the column of free terms: 1,1,1)

det2 = (1-a)(c-a)(c-1)

det3 = (1-a)(1-b)(b-a)

x=(b-1)(c-1)(c-b) / (b-a)(c-a)(c-b) = (b-1)(c-1) / (b-a)(c-a)

y = (1-a)(c-a)(c-1) / (b-a)(c-a)(c-b)

y =(1-a)(c-1) /(b-a)(c-b)

z = (1-a)(1-b)(b-a) / (b-a)(c-a)(c-b)

z = (1-a)(1-b) / (c-a)(c-b)

To find the solutions to the equations:

x+y+z=1......................................(1)

ax+by+cz=1.................................(2)

ax^2+b^2y+c^2z=1....................(3).

Solutions:

Let us eliminate z by (2)-(1)*c and (3)-(1)*c :

(a-c)x+(b-c)y = 1-c.............................(4)

(a^2-c^2) +(b^2-c^2)y = 1-c^2..........(5).

(5) - (1)(b+c) eliminates y:

(a^2-c^2)x - (a-c)(b+c)x = (1-c^2 - (1-c)(b+c)

{a^2-c^2 -ab-ac+bc+c^2}x = 1-c^2 -b-c+bc+c^2

{a^2-ab-ac+bc}x = {1-b-c+bc} = (1-b)(1-c)

(a-b)(a-c)x= (1-b)(1-c)

x = (1-b)(1-c)/(a-b)(a-c).

Similarly it can be shown that:

y = (1-a)(1-c)/(b-a)(b-c)

z = (1-a)(1-b)/(c-a)(c-b)