Question

solve this question immediately

Answer 1

heyyy frnds Ur answer is.....


(a-b)²+ (b-c)²+ (c-a)²

Answer 2

Solution :

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Given :

To find the value of :

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

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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)

Hence,


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

⇒ $\frac{-3(a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2)}{(a-b)(b-c)(a-c)}$

⇒ $\frac{-3ab}{(b-c)(c-a)} + \frac{-3bc}{(a-b)(c-a)} + \frac{-3ac}{(a-b)(b-c)}$

⇒ $-3( \frac{ab}{(b-c)(c-a)} + \frac{bc}{(a-b)(c-a)} + \frac{ac}{(a-b)(b-c)} )$

⇒ $\frac{-3(a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2)}{(-a^2b - ab^2 - b^2c - bc^2 - c^2a - ca^2)}$

⇒ $\frac{-3(a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2)}{-((a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2))}$

⇒ 3,.

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