Question

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

Answer 1

$\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|$

$c_1\to c_1-c_2,c_2\to c_2-c_3$

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

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

$\left\{\right(A^3-B^3)=(A-B\left)\right(A^2+B^2+AB\left)\right\}$

taking $(a-b)$ from $c_1$ and $(b-c)$ from $c_2$

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

$\Rightarrow (a-b)(b-c)[0-0+(b^2+c^2+bc-a^2-b^2-an\left)\right]$

$\Rightarrow (a-b)(b-c)[c^2-a^2+b(c-a\left)\right]$

$\Rightarrow (a-b)(b-c)\left[\right(c-a\left)\right(c+a)+b(c-a\left)\right]$

$\Rightarrow (a-b)(b-c)(c-a)(a+b+c)$