Question

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

• A. $a-b-c$
• B. $a-b+c$
• C. $a+b+c$
• D. $0$

Answer 1

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

Expressing the given determinant as sum of two determinants, we get
$\left|\begin{array}{ccc}a& a^3& a^4\\ b& b^3& b^4\\ c& c^3& c^4\end{array}\right|-\left|\begin{array}{ccc}a& a^3& 1\\ b& b^3& 1\\ c& c^3& 1\end{array}\right|=0,$
We take a,b,c common from the first determinant and apply $R_2\to R_2-R_1,R_3\to R_3-R_1$ in both determinants.
$\Rightarrow abc\left|\begin{array}{ccc}1& a^2& a^3\\ 0& b^2-a^2& b^3-a^3\\ 0& c^2-a^2& c^3-a^3\end{array}\right|=\left|\begin{array}{ccc}a& a^3& 1\\ b-a& b^3-a^3& 0\\ c-a& c^3-a^3& 0\end{array}\right|$

As $a,b,c$ are all distinct, canceling out $b-a$ and $c-a$, we get

$abc\left|\begin{array}{cc}b+a& b^2+a^2+ab\\ c+a& c^2+a^2+ac\end{array}\right|=\left|\begin{array}{cc}1& b^2+a^2+ab\\ 1& c^2+a^2+ac\end{array}\right|$
Applying $R_2\to R_2-R_1$ and then canceling $c-b$ on both sides, we get

$abc\left|\begin{array}{cc}b+a& b^2+a^2+ab\\ 1& a+b+c\end{array}\right|=\left|\begin{array}{cc}1& b^2+a^2+ab\\ 0& a+b+c\end{array}\right|$

$\therefore abc(ab+b^2+bc+a^2+ab+ac-b^2-a^2-ab)=a+b+c$
$\Rightarrow abc(ab+bc+ca)=a+b+c$