Question

If the centroid of the triangle formed by the points (a, b), (1, a) and (b, 1) is at the origin, then $\frac{a^3+b^3+1}{ab}=$ ____________.

Answer 1

$$\begin{gathered} \mathrm{Centroid}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{with}\mathrm{vertices}\left(x_1,y_1\right),\left(x_2,y_2\right)\mathrm{and}\left(x_3,y_3\right)\mathrm{is}\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right). \\ \mathrm{Centroid}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{with}\mathrm{vertices}\left(a,b\right),\left(1,a\right)\mathrm{and}\left(b,1\right)=\left(0,0\right) \\ \Rightarrow \left(\frac{a+1+b}{3},\frac{b+a+1}{3}\right)=\left(0,0\right) \\ \Rightarrow \frac{a+b+1}{3}=0 \\ \Rightarrow a+b+1=0 \\ \Rightarrow a+b=-1 \\ \mathrm{Now}, \\ \frac{a^3+b^3+1}{ab}=\frac{{\left(a+b\right)}^3-3ab\left(a+b\right)+1}{ab} \\ =\frac{{\left(-1\right)}^3-3ab\left(-1\right)+1}{ab}\left(\because a+b=-1\right) \\ =\frac{-1+3ab+1}{ab} \\ =\frac{3ab}{ab} \\ =3 \end{gathered}$$

Hence, $\frac{a^3+b^3+1}{ab}=$3.