If the straight lines ax +by+c=0, bx +cy +a = 0 and cx +ay + b = 0 are concurrent, then prove that ${\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3=3\mathrm{abc}$

Given st.lines
$\begin{array}{l}\mathrm{ax}+\mathrm{by}+\mathrm{c}=0 ....\left(1\right)\\ \mathrm{bx}+\mathrm{cy}+\mathrm{a}=0 ...\left(2\right)\end{array}$

Now point of intersection of (1) and (2) is

$$\begin{gathered} x y 1 \\ b c a b \\ c a b c \end{gathered}$$

$\begin{array}{l}\frac{\mathrm{x}}{\mathrm{ba}-{\mathrm{c}}^2}=\frac{\mathrm{y}}{\mathrm{bc}-{\mathrm{a}}^2}=\frac{1}{\mathrm{ac}-{\mathrm{b}}^2}\\ (\mathrm{x},\mathrm{y})=\left(\frac{\mathrm{ab}-{\mathrm{c}}^2}{\mathrm{ac}-{\mathrm{b}}^2},\frac{\mathrm{bc}-{\mathrm{a}}^2}{\mathrm{ac}-{\mathrm{b}}^2}\right)\end{array}$
If the lines L₁, L₂, L₃ are concurrent, L₃ contains the above point of intersection of L₁ and L₂
$\begin{array}{l}\therefore \mathrm{c}\left(\frac{\mathrm{ab}-{\mathrm{c}}^2}{\mathrm{ac}-{\mathrm{b}}^2}\right)+\mathrm{a}\left(\frac{\mathrm{bc}-{\mathrm{a}}^2}{\mathrm{ca}-{\mathrm{b}}^2}\right)+\mathrm{b}=0\\ \Rightarrow \mathrm{c}\left(\mathrm{ab}-{\mathrm{c}}^2\right)+\mathrm{a}\left(\mathrm{bc}-{\mathrm{a}}^2\right)+\mathrm{b}\left(\mathrm{ca}-{\mathrm{b}}^2\right)=0\\ \Rightarrow {\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3=3\mathrm{abc}\end{array}$