Math, asked by rachnakaur220683, 11 months ago

find
the value of (x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c);if.a +b +c=3x​

Answers

Answered by mfb8525
2

Answer:

We have,  

(x-a)3 +(x-b)3 +(x-c)3 -3(x-a) (x-b) (x-c) and a + b + c = 3x .... (1)

We know that if p + q + r = 0 then  p3 + q3 + r3 = 3 pqr

Here, p + q + r = (x-a) + (x-b) + (x-c) = 3x - a - b - c

3x - (a + b + c) = 3x - 3x  [Using (1)]

                         = 0

So, by definition,

(x-a)3 +(x-b)3 +(x-c)3 = 3(x-a) (x-b) (x-c)

that implies,

(x-a)3 +(x-b)3 +(x-c)3 - 3(x-a) (x-b) (x-c) = 0

hence your ans is 0.

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