Question

If the centroid of a triangle formed by the points \((a,b), (b, c)\) and \((c,a)\) is at the origin, then \(a^3+b^3+c^3 =\)

• A. abc
• B. 0
• C. a+b+c
• D. 3abc

Answer 1

The Correct Option is D

The centroid of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by $$(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$$ In this case, the vertices are (a, b), (b, c), and (c, a). The centroid is at the origin (0, 0). Thus, $$\frac{a+b+c}{3} = 0$$ $$\frac{b+c+a}{3} = 0$$ From both equations, we get $$a+b+c = 0$$ We know the identity: $$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)$$ Since a + b + c = 0, we have $$a^3 + b^3 + c^3 - 3abc = 0$$ $$a^3 + b^3 + c^3 = 3abc$$