Question

1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix} \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. None of these

Answer 1

Answer: (C)

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

Answer 2

$\Delta = \left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix} \right|$ vanishes when $a = b,b = c,c = a$. Hence $(a - b),(b - c),(c - a)$ are factors of $\Delta$. Since $\Delta$ is symmetric in $C_{21} = ( - 1)^{2 + 1}(18 + 21) = - 39$ and of 4^th degree, $(a + b + c)$ is also a factor, so that we can write

$- 10x + 90 - 42 - 81 + 42 + 9x = 0\text{or}\ x = 9$ ......(i)

Where by comparing the coefficients of the leading term $bc^{3}$ on both the sides of identity (i). We get

$1 = k( - 1)( - 1) \Rightarrow k = 1$

$C_{3} \rightarrow C_{3} - C_{1},$.

Trick : Put $a = 1,b = 2,c = 3$, so that determinant

$\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 8 & 27 \end{matrix} \right| = 1(30) - 1(24) + 1(8 - 2) = 12$ which is given by (3). i.e. $(1 + 2 + 3)(1 - 2)(2 - 3)(3 - 1) = 12$.