$∣ ∣ ∣ ∣ \begin{matrix} b + c & a - b & a c + a & b - c & b a + b & c - a & c \end{matrix} ∣ ∣ ∣ ∣ =$

a^{3} + b^{3} + c^{3} - 3 a b c

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3 a b c - a - b^{3} - c^{3}

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3 a b c

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Solution

The correct option is D $3 a b c - a - b^{3} - c^{3}$
Performing $C_{2} \to C_{2} - C_{3}$ in the given equation, we get
$∣ ∣ ∣ ∣ \begin{matrix} b + c & - b & a c + a & - c & b a + b & - a & c \end{matrix} ∣ ∣ ∣ ∣$
Performing $C_{1} \to C_{1} + C_{2}$
$∣ ∣ ∣ ∣ \begin{matrix} c & - b & a a & - c & b b & - a & c \end{matrix} ∣ ∣ ∣ ∣$
$= c (- c^{2} + a b) + b (a c - b^{2}) + a (- a^{2} + b c)$
$= - c^{3} + a b c + b a - + b^{3} - a^{3} + a b c$
$= 3 a b c - a^{3} - b^{3} - c^{3}$

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Similar questions

If a + b + c = 9 and ab + bc + ca = 23, then $a^{3} + b^{3} + c^{3} - 3 a b c =$

∣ ∣ ∣ ∣ \begin{matrix} b + c & a - b & a c + a & b - c & b a + b & c - a & c \end{matrix} ∣ ∣ ∣ ∣ = 3 a b c - a^{3} - b^{3} - c^{3}

a + b + c = 9

ab + bc + ca = 26,

a^{3} + b^{3} + c^{3} - 3abc

\frac{a^{3} + b^{3} + c^{3} - 3 a b c}{a^{2} + b^{2} + c^{2} - a b - b c - c a}

⎛ ⎜ ⎝ \begin{matrix} b + c & a - b & a c + a & b - c & b a + b & c - a & c \end{matrix} ⎞ ⎟ ⎠

3 a b c - a^{3} - b^{3} - c^{3}