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Differentiation to Solve Modified Sum of Binomial Coefficients

Prove that:b+...

Question

Prove that :

$|\begin{matrix} {(b + c)}^{2} & a^{2} & b c \\ {(c + a)}^{2} & b^{2} & c a \\ {(a + b)}^{2} & c^{2} & a b \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2})$

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Solution

$Let LHS = Δ = |{(b + c)}^{2} a^{2} b c {(c + a)}^{2} b^{2} c a {(a + b)}^{2} c^{2} a b| = |{(b + c)}^{2} - {(c + a)}^{2} a^{2} - b^{2} bc - ca {(c + a)}^{2} - {(a + b)}^{2} b^{2} - c^{2} c a - a b {(a + b)}^{2} c^{2} a b| [Applying R_{1} \to R_{1} - R_{2} and R_{2} \to R_{2} - R_{3}] = |(b - a) (b + 2 c + a) (a + b) (a - b) c (b - a) (c - b) (b + 2 a + c) (b - c) (b + c) a (c - b) {(a + b)}^{2} c^{2} a b| = (a - b) (b - c) |- (b + 2 c + a) a + b - c - (b + 2 a + c) b + c - a {(a + b)}^{2} c^{2} a b| [A pplying x^{2} - y^{2} = (x + y) (x - y) and taking out (a - b) co m m o n from R_{1} and (b - c) from R_{2}] = (a - b) (b - c) |- 2 (b + c + a) a + b - c - 2 (b + a + c) b + c - a {(a + b)}^{2} - c^{2} c^{2} a b| [Applying C_{1} \to C_{1} - C_{2}] = (a - b) (b - c) |- 2 (b + c + a) a + b - c - 2 (b + a + c) b + c - a (a + b + c) (a + b - c) c^{2} a b| [Applying x^{2} - y^{2} = (x + y) (x - y) in C_{1}] = (a - b) (b - c) (a + b + c) |- 2 a + b - c - 2 b + c - a (a + b - c) c^{2} a b| [Taking out (a + b + c) common from C_{1}] = (a - b) (b - c) (a + b + c) |- 2 a + b - c 0 c - a c - a (a + b - c) c^{2} a b| [Applying R_{2} \to R_{2} - R_{1}] = (a - b) (b - c) (a + b + c) (c - a) |- 2 a + b - c 0 1 1 (a + b - c) c^{2} a b| [Taking out (c - a) common from R_{2}] = (a - b) (b - c) (a + b + c) (c - a) |- 2 a + b + c - c 0 0 1 (a + b - c) c^{2} - ab a b| [Applying C_{2} \to C_{2} - C_{3}] = (a - b) (b - c) (a + b + c) (c - a) \{(- 1) |- 2 a + b + c (a + b - c) c^{2} - ab|\} [Expanding along R_{2}] = - (a - b) (b - c) (a + b + c) (c - a) \{- 2 c^{2} + 2 ab - a^{2} - b^{2} - 2 ab + c^{2}\} = - (a - b) (b - c) (a + b + c) (c - a) (- a^{2} - b^{2} - c^{2}) = (a - b) (b - c) (a + b + c) (c - a) (a^{2} + b^{2} + c^{2}) = RHS$

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Similar questions

Q. Prove the following :

∣ ∣ ∣ ∣ \begin{matrix} a & b - c & c + b a + c & b & c - a a - b & b + a & c \end{matrix} ∣ ∣ ∣ ∣ = (a + b + c) (a^{2} + b^{2} + c^{2}) .

Q. Using properties of determinants, prove that :

∣ ∣ ∣ ∣ \begin{matrix} a & b - c & c + b a + c & b & c - a a - b & b + a & c \end{matrix} ∣ ∣ ∣ ∣ = (a + b + c) (a^{2} + b^{2} + c^{2})

∣ ∣ ∣ ∣ \begin{matrix} a & b - c & c + b a + c & b & c - a a - b & b + a & c \end{matrix} ∣ ∣ ∣ ∣ = (a + b + c) (a^{2} + b^{2} + c^{2})

Q. Prove that :

|\begin{matrix} a^{2} & a^{2} - {(b - c)}^{2} & b c \\ b^{2} & b^{2} - {(c - a)}^{2} & c a \\ c^{2} & c^{2} - {(a - b)}^{2} & a b \end{matrix}| = (a - b) (b - c) (c - a) (a + b + c) (a^{2} + b^{2} + c^{2})

Q. (a − b)³ + (b − c)³ + (c − a)³ =

(a) (a + b + c) (a² + b²+ c² − ab − bc − ca)

(b) (a − b) (b − c) (c − a)

(c) 3(a − b) ( b− c) (c − a)

(d) none of these

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