Question

x + y + z + 1 = 0
ax + by + cz + d = 0
a²x + b²y + x²z + d² = 0

Answer 1

These equations can be written as 
\[x + y + z = - 1\] 
\[ax + by + cz = - d\] 
\[ a^2 x + b^2 y + x^2 z = - d^2 \] 
\[D = \begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{vmatrix} \] 
\[ = \begin{vmatrix}1 & 0 & 0 \\ a & a - b & b - c \\ a^2 & a^2 - b^2 & b^2 - c^2\end{vmatrix} \left[\text{ Applying }C_2 \to C_1 - C_2 , C_3 \to C_2 - C_3 \right]\] 
\[\text{Taking (b - a) and (c - a) common from }C_1\text{ and }C_2 ,\text{ respectively, we get }\] 
\[ = (a - b)(b - c)\begin{vmatrix}1 & 0 & 0 \\ a & 1 & 1 \\ a^2 & a + b & b + c\end{vmatrix}\] 
\[ = (a - b)(b - c)(c - a) \ldots(1)\] 
\[ D_1 = \begin{vmatrix}- 1 & 1 & 1 \\ - d & b & c \\ - d^2 & b^2 & c^2\end{vmatrix} = - \begin{vmatrix}1 & 1 & 1 \\ d & b & c \\ d^2 & b^2 & c^2\end{vmatrix}\] 
\[ D_1 = - (d - b) (b - c) (c - d) \left[\text{ Replacing a by d in eq }. (1) \right]\] 
\[ D_2 = \begin{vmatrix}1 & - 1 & 1 \\ a & - d & c \\ a^2 & - d^2 & c^2\end{vmatrix} = - \begin{vmatrix}1 & 1 & 1 \\ a & d & c \\ a^2 & d^2 & c^2\end{vmatrix}\] 
\[ D_2 = - (a - d)(d - c)(c - a) \left[\text{ Replacing b by d in eq }. (1) \right]\] 
\[ D_3 = \begin{vmatrix}1 & 1 & - 1 \\ a & b & - d \\ a^2 & b^2 & - d^2\end{vmatrix} = - \begin{vmatrix}1 & 1 & 1 \\ a & b & d \\ a^2 & b^2 & d^2\end{vmatrix}\] 
\[ D_3 = - (a - b)(b - d)(d - a) \left[\text{ Replacing c by d in eq }. (1) \right]\] 
Thus,
\[x = \frac{D_1}{D} = - \frac{(d - b)(b - c)(c - d)}{(a - b)(b - c)(c - a)}\] 
\[y = \frac{D_2}{D} = - \frac{(a - d)(d - c)(c - a)}{(a - b)(b - c)(c - a)}\] 
\[z = \frac{D_3}{D} = - \frac{(a - b)(b - d)(d - a)}{(a - b)(b - c)(c - a)}\]