Question

If the point $\left(a^3\right(a-1),a^2-3(a-1\left)\right),\left(b^3\right(b-1),b^2-3(b-1\left)\right)$ and $\left(\frac{c^3}{(}c-1\right),c^2-3\left(c-1\right))$ are collinear for three distinct values $a,b,c$ and $a\ne 1,b\ne 1$ and $c\ne 1$, then find the value of $abc-(ab+bc+ac)+3(a+b+c)$.

Answer 1

$\left|\begin{array}{ccc}{a}^3\left(a-1\right)& a^2-3\left(a-1\right)& 1\\ b^3\left(b-1\right)& b^2-3\left(b-1\right)& 1\\ c^3\left(c-1\right)& c^2-3\left(c-1\right)& 1\end{array}\right|$ where $a,b,c\ne 1$

$=(a-1)(b-1)(c-1)\left|\begin{array}{ccc}{a}^3& a^2-3& a-1\\ b^3& b^2-3& b-1\\ c^3& c^2-3& c-1\end{array}\right|=0$

$R_1\to R_1-R_2$ and $R_2\to R_2-R_3$

$=(a-1)(b-1)(c-1)\left|\begin{array}{ccc}{a}^3-b^3& a^2-b^2& a-b\\ b^3-c^3& b^2-c^2& b-c\\ c^3& c^2-3& c-1\end{array}\right|=0$

$=(a-1)(b-1)(c-1)(a-b)(b-c)\left|\begin{array}{ccc}{a}^2+ab+b^2& a+b& 1\\ b^2+bc+c^2& b+c& 1\\ c^3& c^2-3& c-1\end{array}\right|=0$

$C_1\to C_1-C_2$

$=(a-1)(b-1)(c-1)(a-b)(b-c)\left|\begin{array}{ccc}{a}^2+ab-bc-c^2& a-c& 0\\ b^2+bc+c^2& b+c& 1\\ c^3& c^2-3& c-1\end{array}\right|=0$

$=(a-1)(b-1)(c-1)(a-b)(b-c)\left|\begin{array}{ccc}(a+b+c)(a-c)& a-c& 0\\ b^2+bc+c^2& b+c& 1\\ c^3& c^2-3& c-1\end{array}\right|=0$

$=(a-1)(b-1)(c-1)(a-b)(b-c)(a-c)\left|\begin{array}{ccc}a+b+c& 1& 0\\ b^2+bc+c^2& b+c& 1\\ c^3& c^2-3& c-1\end{array}\right|=0$

$=(a-1)(b-1)(c-1)(a-b)(b-c)(a-c)[(a+b+c)(bc-b+c^2-c-c^2+3)-1(b^2{c}^2-3b^2+b{c}^3-3bc+c^4-3c^2-b{c}^3-c^4)]$

$=abc-(ab+bc+ca)+3(a+b+c)=0$ on simplification