Question

If a,b,c are distinct and $\left|\begin{array}{ccc}a& a^2& a^3-1\\ b& b^2& b^3-1\\ c& c^2& c^3-1\end{array}\right|=0$ then

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

Answer 1

The correct option is D $abc=1$

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

$\Rightarrow a(b^2{c}^3-b^2-c^2{b}^3+c^2)-a^2(b{c}^3-b-c{b}^3+c)+(a^3-1)(b{c}^2-c{b}^2)=0$

$\Rightarrow a\left(b^2{c}^2\right(c-b)+c^2-b^2)-a^2\left(bc\right(c^2-b^2)+c-b)+(a^3-1)bc(c-b)=0$

$\Rightarrow a\left(b^2{c}^2\right(c-b)+(c-b\left)\right(c+b\left)\right)-a^2\left(bc\right(c-b\left)\right(c+b)+(c-b\left)\right)+bc(a^3-1)(c-b)=0$

$\Rightarrow a(c-b)(b^2{c}^2+c+b)-a^2(c-b)\left(bc\right(c+b)+1)+bc(a^3-1)(c-b)=0$

$\Rightarrow (c-b)\left\{a\right(b^2{c}^2+c+b)-a^2(bc(c+b)+1)+bc(a^3-1\left)\right\}=0$

$\Rightarrow (c-b)\{a{b}^2{c}^2+ac+ab-a^{2}b{c}^2-a^2{b}^{2}c-a^2+a^{3}bc-bc\}=0$

$\Rightarrow (c-b)\left\{abc\right(bc-ac-ab+b^2)-(bc-ac-ab+a^2\left)\right\}=0$

$\Rightarrow (c-b)(bc-ac-ab+a^2)(abc-1)=0$

$\Rightarrow (c-b)\left(b\right(c-a)-a(c-a\left)\right)(abc-1)=0$

$\Rightarrow (c-b)(c-a)(b-a)(abc-1)=0$

$\Rightarrow c-b=0$ or $c-a=0$ or $b-a=0$

$\Rightarrow c=b$ $c=a$ $b=a$

which is not possible since $a,b,c$ are distinct.

$\therefore abc-1=0$

$\Rightarrow abc=1$