Question

If $\left|\begin{array}{ccc}a& a^3& a^4-1\\ b& b^3& b^4-1\\ c& c^3& c^4-1\end{array}\right|=0$ and $a,b,c$ are all distinct, then $abc(ab+bc+ca)$ is equal to

• A. $a+b+c$
• B. $abc$
• C. $0$
• D. none of these

Answer 1

The correct option is A $a+b+c$

Given $\left|\begin{array}{ccc}a& a^3& a^4-1\\ b& b^3& b^4-1\\ c& c^3& c^4-1\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}a& a^3& a^4\\ b& b^3& b^4\\ c& c^3& c^4\end{array}\right|+\left|\begin{array}{ccc}a& a^3& -1\\ b& b^3& -1\\ c& c^3& -1\end{array}\right|=0$

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

$\Rightarrow abc(a-b)(b-c)(c-a)(ab+bc+ca)-(a-b)(b-c)(c-a)(a+b+c)=0$

$\Rightarrow abc(ab+bc+ca)-(a+b+c)=0$

$\Rightarrow abc(ab+bc+ca)=a+b+c$