Question

If $\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=k(a-b)(b-c)(c-a)$, then $k$ is equal to

• A. $-2$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

Answer: (B)

$1$

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

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

$=bc(c-b)-a(c-b)(c+b)+a^2(c-b)$

$=(c-b)[bc-a(c+b)+a^2]$

$=(c-b)[bc-ac-ab+a^2]$

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

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

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

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

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

ANSWER IS B