Question

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

• A. $a^3+b^3+c^3-3abc$
• B. $a^3+b^3+c^3+3abc$
• C. $(a+b+c)(a-b)(b-c)(c-a)$
• D. $Noneofthese$

Answer 1

The correct option is C $(a+b+c)(a-b)(b-c)(c-a)$

$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^3& b^3& c^3\end{array}\right|$ vanishes when a=b, b=c,c=a. Hence (a-b),(b-c),(c-a) are factors of $\Delta$. Since $\Delta$ is symmetric in a,b,c adn of
$4^{th}$ degree, (a+b+c) is also a factor, so that we can write
$\Delta =k(a-b)(b-c)(c-a)(a+b+c)$ ...(i)
Where by comparing the coefficients of the leading term $b{c}^3$ on both the sides of identity (i). We get
$1=k(-1)(-1)\Rightarrow k=1$
$\therefore \Delta =(a-b)(b-c)(c-a)(a+b+c)$.
Trick : Put a=1,b=2,c=3, so that determinant $\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& 3\\ 1& 8& 27\end{array}\right|=1\left(30\right)-1\left(24\right)+1(8-2)=12$ which is given by (c).
i.e., (1+2+3)(1-2)(2-3)(3-1)=12.