Question

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

• A. abc
• B. 0.0
• C. a+b+c
• D. 3abc

Answer 1

The correct option is D 3abc

Given centroid is (0, 0).
$\because$ The centroid of triangle formed by $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ is $\left(\frac{x_1+x_2+x_3}{3}\right),\left(\frac{y_1+y_2+y_3}{3}\right)$

$\Rightarrow \frac{a+b+c}{3}=0$

$\Rightarrow a+b+c=0$..........(i)
$\Rightarrow a+b=-c,b+c=-a,a+b=-c$.....(ii)

We know that:

$\Rightarrow (a+b+c{)}^3=a^3+b^3+c^3+3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc$

$\Rightarrow a^3+b^3+c^3=(a+b+c{)}^3-[3a^2(b+c)+3b^2(a+c)+3c^2(a+b)+6abc]$

Substituting (i) and (ii) in the above equation we get,

$\Rightarrow a^3+b^3+c^3=-3a^2(-a)-3b^2(-b)-3c^2(-c)-6abc]$

$\Rightarrow a^3+b^3+c^3=3a^3+3b^3+3c^3-6abc$

$\Rightarrow -2a^3-2b^3-2c^3=-6abc$

$\therefore a^3+b^3+c^3=3abc$