Using properties of determinants prove the following:
$|\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c)$

Solution

$∆ = |\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix}| Applying C_{1} \to C_{1} - C_{3} and C_{2} \to C_{2} - C_{3}, we have : ∆ = |\begin{matrix} 1 - 1 & 1 - 1 & 1 \\ a - c & b - c & c \\ a^{3} - c^{3} & b^{3} - c^{3} & c^{3} \end{matrix}| = |\begin{matrix} 0 & 0 & 1 \\ a - c & b - c & c \\ (a - c) (a^{2} + a c + c^{2}) & (b - c) (b^{2} + b c + c^{2}) & c^{3} \end{matrix}| = (c - a) (b - c) |\begin{matrix} 0 & 0 & 1 \\ - 1 & 1 & c \\ - a^{2} + a c + c^{2} & b^{2} + b c + c^{2} & c^{3} \end{matrix}|$

$Applying C_{1} \to C_{1} + C_{2}, we have : ∆ = (c - a) (b - c) |\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & c \\ b^{2} - a^{2} + b c - a c & b^{2} + b c + c^{2} & c^{3} \end{matrix}| = (b - c) (c - a) (a - b) |\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & c \\ - a + b + c & b^{2} + b c + c^{2} & c^{3} \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c) |\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & c \\ - 1 & b^{2} + b c + c^{2} & c^{3} \end{matrix}|$

Expanding along C₁, we have:

$∆ = (a - b) (b - c) (c - a) (a + b + c) - 1 |\begin{matrix} 0 & 1 \\ 1 & c \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c)$

Hence proved.

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Relations And Functions

Using properties of determinants prove the following:
$|\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c)$

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Relations And Functions

Using properties of determinants prove the following: 1 1 1a b ca3 b3 c3 = a - b b - c c - a a + b +c

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Mock Test Series

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Using properties of determinants prove the following:
$|\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c)$