Question

If the centroid of triangle A, B, C whose A(a,b), B(b,c), C(c,a) lies at origin. Calculate$\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}}$.

Answer 1

Hint:
The centroid is the center point of the object. The point in which the three medians of the triangle intersect is known as the centroid of a triangle. The medians are the segments that connect a vertex to the midpoint of the opposite side. It can be found by taking the average of x-coordinate points and y-coordinate points of all the vertices of the triangle. If the three vertices of the triangles are A\[\left( {{x_1},{y_1}} \right)\], B\[\left( {{x_2},{y_2}} \right)\] and C\[\left( {{x_3},{y_3}} \right)\] then the centroid is given by:
Centroid (G) $ = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$
Where ${x_1},{x_2},{x_3}$are the x-coordinates of the vertices.
${y_1},{y_2},{y_3}$ are the y-coordinates of the vertices.
The centroid is also called the center of gravity of the triangle.
If $a + b + c = 0$ then ${a^3} + {b^3} + {c^3} = 3abc$

 Complete step by step solution:
The point of concurrence of medians is called its centroid.
Given that the centroid of the triangle is origin.

Now, Centroid of ∆ABC is given by,
$G = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$
$
  \left( {0,0} \right) = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right) \\
  \left( {0,0} \right) = \left( {\dfrac{{a + b + c}}{3},\dfrac{{b + c + a}}{3}} \right) \\
\dfrac{{a + b + c}}{3} = 0 \\
a + b + c = 0 \ldots (1) \\
$
Now, consider the identity,
${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$
If $a + b + c = 0$ then
$
{a^3} + {b^3} + {c^3} - 3abc = 0 \\
{a^3} + {b^3} + {c^3} = 3abc \ldots (2) \\
$
Now, by question
$\dfrac{{{a^3} + {b^3} + {c^3}}}{{abc}} = \dfrac{{3abc}}{{abc}}$ = 3

 Note: The centroid of the triangle divides the median in ratio 2:1 internally. The distance from the centroid to the vertex is twice as long as the distance from the centroid to the midpoint of the side opposite the vertex. Two points coincide only when their abscissa as well as ordinates is equal.
The centroid of the triangle is at two-third of the distance from the vertex to the midpoint of the sides.
 Centroid always lies inside the object and it is the point of concurrency of the medians. It is also called the center of gravity of the triangle.
Also whenever $a + b + c = 0$, \[{a^3} + {b^3} + {c^3} = 3abc\]