Question

If the straight line ax+by+c=0 ,bx+cy+a=0 and cx+ay+b=0 are concurrent then prove that a^3+b^3+c^3=3abc

Answer 1

Answer:

The three lines are concurrent

=> there exists a pair (x,y) that makes the 3 equations hold

=> there exists a solution to the matrix equation:

    $\displaystyle\left(\begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array}\right)\left(\begin{array}{c}x\\y\\1\end{array}\right)=\left(\begin{array}{c}0\\0\\0\end{array}\right)$

=> the determinant of the coefficient matrix is zero:

$\Rightarrow\left|\begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array}\right|=0\\\\\Rightarrow 3abc-a^3-b^3-c^3=0\\\\\Rightarrow a^3+b^3+c^3 = 3abc$