# Solve this

Question:

$x+y+z+1=0$

$a x+b y+c z+d=0$

$a^{2} x+b^{2} y+x^{2} z+d^{2}=0$

Solution:

These equations can be written as

$x+y+z=-1$

$a x+b y+c z=-d$

$a^{2} x+b^{2} y+x^{2} z=-d^{2}$

$D=\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2}\end{array}\right|$

$=\left|\begin{array}{ccc}1 & 0 & 0 \\ a & a-b & b-c \\ a^{2} & a^{2}-b^{2} & b^{2}-c^{2}\end{array}\right|$          [Applying $C_{2} \rightarrow C_{1}-C_{2}, C_{3} \rightarrow C_{2}-C_{3}$ ]

Taking $(b-a)$ and $(c-a)$ common from $C_{1}$ and $C_{2}$, respectively, we get

$=(a-b)(b-c)\left|\begin{array}{ccc}1 & 0 & 0 \\ a & 1 & 1 \\ a^{2} & a+b & b+c\end{array}\right|$

$=(a-b)(b-c)(c-a) \quad \ldots(1)$

$D_{1}=\left|\begin{array}{lll}-1 & 1 & 1 \\ -d & b & c \\ -d^{2} & b^{2} & c^{2}\end{array}\right|=-\left|\begin{array}{ccc}1 & 1 & 1 \\ d & b & c \\ d^{2} & b^{2} & c^{2}\end{array}\right|$

$D_{1}=-(d-b)(b-c)(c-d) \quad$ [Replacing $a$ by $d$ in eq. (1)]

$D_{2}=\left|\begin{array}{ccc}1 & -1 & 1 \\ a & -d & c \\ a^{2} & -d^{2} & c^{2}\end{array}\right|=-\left|\begin{array}{ccc}1 & 1 & 1 \\ a & d & c \\ a^{2} & d^{2} & c^{2}\end{array}\right|$

$D_{2}=-(a-d)(d-c)(c-a)$  [Replacing $b$ by $d$ in eq. (1)

$D_{3}=\left|\begin{array}{ccc}1 & 1 & -1 \\ a & b & -d \\ a^{2} & b^{2} & -d^{2}\end{array}\right|=-\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & d \\ a^{2} & b^{2} & d^{2}\end{array}\right|$

$D_{3}=-(a-b)(b-d)(d-a)$       [Replacing $c$ by $d$ in eq. (1)]

Thus,

$x=\frac{D_{1}}{D}=-\frac{(d-b)(b-c)(c-d)}{(a-b)(b-c)(c-a)}$

$y=\frac{D_{2}}{D}=-\frac{(a-d)(d-c)(c-a)}{(a-b)(b-c)(c-a)}$

$z=\frac{D_{3}}{D}=-\frac{(a-b)(b-d)(d-a)}{(a-b)(b-c)(c-a)}$