Question

factorise (a-b)³+ (b-c)³+ (c-a)³

Answer 1

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Answer 2

Solution :

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Given :

To find the value of :   (a-b)³+ (b-c)³+ (c-a)³

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We know the identity that,

(a - b)³ = a³ - 3a²b + 3ab² - b³

Hence,

Applying the identity with the values

we get,


(a - b)³ = a³ - 3a²b + 3ab² - b³,

(b - c)³ = b³ - 3b²c + 3bc² - c³,

(c - a)³ = c³ - 3c²a + 3ca² - a³

Hence,

(a - b)³ + (b - c)³ + (c - a)³ = a³ - 3a²b + 3ab² - b³ + b³ - 3b²c + 3bc² - c³ + c³ -3c²a +3ca² - a³

⇒ a³ - a³ + b³ - b³ + c³ - c³ -3a²b + 3ab² -3b²c + 3bc² + 3c²a + 3ca²

⇒ - 3ab(a-b) - 3bc(b - c) - 3ac(c - a)

⇒ -3(a²b + ab² + b²c + bc² + c²a + ca²) (or -3[ab(a-b)+bc(b-c)+ac(a-c)])
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