Question

$det\left[\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right]$ is equal to

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

Answer 1

Answer: (B)

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

Let $A=det\left[\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right]$

So, By row transformation,

$r_1\to r_1-r_2$ and $r_3\to r_3-r_2$

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

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

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

$\Rightarrow A=(a-b)(b-c)(c-a)$