Question

lf $a,b,c$ are distinct and $\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|=0$,then:

• A. $a+b+c=1$
• B. $ab+bc+ca=0$
• C. $a+b+c=0$
• D. $abc=0$

Answer 1

The correct option is D $abc=0$

Given,$A=\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|=0$

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

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

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

Taking $(a-b)$ and $(c-b)$ common from respective rows

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

$=abc(a-b)(c-b)(-1(c+b-a-b\left)\right)$

$A=abc(a-b)(b-c)(c-a)$

So,$A=abc(a-b)(b-c)(c-a)=0$

then $abc=0$