If a, b, c are all different, solve the system of equations `x+y+z=1` `ax+by+cz=lamda` `a^2x+b^2y+c^2z=lamda^2` using Cramer's rule.

Explanation

The system can be put in the matrix form
`[{:(1,1,1),(a,b,c),(a^2,b^2,c^2):}][{:(x),(y),(x):}]=[{:(1),(lamda),(lamda^2):}]`
i.e., AX = B
where `A=[{:(1,1,1),(a,b,c),(a^2,b^2,c^2):}]`
The determinant of the coeffient matrix,
`Delta=[{:(1,1,1),(a,b,c),(a^2,b^2,c^2):}]=(a-b)(b-c)(c-a)ne0`
Hence the system possesses a unique solution given by
`x/Delta_1=y/Delta_2=z/Delta_2=1/Delta`
`Delta_i` being obtained from `Delta` by replacing `i^(th)` column of `Delta` by the elements of column matrix `[{:(1),(lamda),(lamda^2):}]`,
i.e., `x/|{:(1,1,1),(lamda,b,c),(lamda^2,b^2,c^2):}|=y/|{:(1,1,1),(a,lamda,c),(a^2,lamda^2,c^2):}|=z/|{:(1,1,1),(a,b,lamda),(a^2,b^2,lamda^2):}|=1/|{:(1,1,1),(a,b,c),(a^2,b^2,c^2):}|`
`x=|{:(1,1,1),(lamda,b,c),(lamda^2,b^2,c^2):}|/|{:(1,1,1),(lamda,b,c),(a^2,b^2,c^2):}|=((lamda-b)(b-c)(c-lamda))/((a-b)(b-c)(c-a))=((lamda-c)(c-lamda))/((a-b)(c-a))`
Similarly `((a-lamda)(lamda-c))/((a-b)(b-c))" and z"=((b-lamda)(lamda-a))/((b-c)(c-a))`