Show that the centroid of the triangle with vertices $(a, b-c)$,$(b, c-a)$,and $(c, a-b)$ lies on the $X$-axis.

(N/A) The coordinates of the vertices of the triangle are $(x_1, y_1) = (a, b-c)$,$(x_2, y_2) = (b, c-a)$,and $(x_3, y_3) = (c, a-b)$.
The formula for the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by:
$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$.
Substituting the given values into the formula:
$x$-coordinate $= \frac{a + b + c}{3}$
$y$-coordinate $= \frac{(b-c) + (c-a) + (a-b)}{3}$
Simplifying the $y$-coordinate:
$y$-coordinate $= \frac{b - c + c - a + a - b}{3} = \frac{0}{3} = 0$.
Since the $y$-coordinate of the centroid is $0$,the centroid lies on the $X$-axis.