Factorise ${(b - c)}^{7} + {(c - a)}^{7} + {(a - b)}^{7}$

7 (b - c) (c - a) (a - b) {(a^{2} + b^{2} + c^{2} + b c + c a + a b)}^{2}

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7 (b - c) (c - a) (a - b) {(a^{2} + b^{2} + c^{2} - b c - c a - a b)}^{2}

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7 (b + c) (c + a) (a + b) {(a^{2} + b^{2} + c^{2} - b c - c a - a b)}^{2}

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7 (b + c) (c + a) (a + b) {(a^{2} + b^{2} + c^{2} + b c + c a + a b)}^{2}

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Solution

The correct option is B $7 (b - c) (c - a) (a - b) {(a^{2} + b^{2} + c^{2} - b c - c a - a b)}^{2}$ .

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Similar questions

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

If a + b + c = 0, then prove the following

(a) (b + c) (b − c) + a(a + 2b) = 0

(b) a(a² − bc) + b(b² − ca) + c(c² − ab) = 0

(d)

\frac{a {(b - c)}^{2}}{(c - a) (a - b)} + \frac{b {(c - a)}^{2}}{(a - b) (b - c)} + \frac{c {(a - b)}^{2}}{(b - c) (c - a)} = a + b + c

a, b, c

A, B, C \in R - (0}

a A + b B + c C + \sqrt{(a^{2} + b^{2} + c^{2}) (A^{2} + B^{2} + C^{2})} = 0

\frac{a B}{b A} + \frac{b C}{c B} + \frac{c A}{a C}

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + a b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3. (b - c) (c - a) (a - b) (a + b + c) (a b + b c + c a)

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