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

Solution

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

Expanding along C₁, We have,

$∆ = (a + b + c) ((b - c) (a + b - 2 c) - (c - a) (c + a - 2 b)) \Rightarrow ∆ = (a + b + c) ((ba + b^{2} - 2 bc - ca - cb + 2 c^{2}) (c^{2} + ac - 2 bc - ac - a^{2} + 2 ab)) \Rightarrow ∆ = (a + b + c) (a^{2} + b^{2} + c^{2} - ca - bc - ab) \Rightarrow ∆ = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - bc - ac) \Rightarrow ∆ = a^{3} + b^{3} + c^{3} - 3 abc = R . H . S .$

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Determinants

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

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Determinants

Using properties of determinants prove the following:a bca - b b - c c - ab + c c + a a + b = a3 + b3 + c3 - 3abc

Some More Questions From Determinants Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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