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Algebraic Identities

a-b3+b-c3+c-a...

Question

$(a - b)^{3} + (b - c)^{3} + (c - a)^{3} =$

A

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

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B

$(a - b) (b - c) (c - a)$

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C

$3 (a - b) (b - c) (c - a)$

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D

none of these

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Solution

The correct option is C
$3 (a - b) (b - c) (c - a)$

$∵ (a - b) + (b - c) + (c - a) = 0 ∴ (a - b)^{3} + (b - c)^{3} + (c - a)^{3} = 3 (a - b) (b - c) (c - a)$

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Similar questions

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

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

a^{3} {(b - c)}^{3} + b^{3} {(c - a)}^{3} + c^{3} {(a - b)}^{3} =

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

Q. The value of determinant

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

> 0

, then prove that

\frac{a^{3}}{b^{3}} + \frac{b^{3}}{c^{3}} + \frac{c^{3}}{a^{3}} \geq 3

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