If 3x = a + B + C, Then the Value of (X − A)3 + (X −B)3 + (X − C)3 − 3(X − A) (X − B) (X −C) is - Mathematics

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MCQ

If 3x = a + b + c, then the value of (x − a)3 + (x −b)3 + (x − c)3 − 3(x − a) (x − b) (x −c) is

Options

  • a + b + c

  • (a − b) (b − c) (c − a)

  • 0

  • none of these

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Solution

The given expression is

 (x − a)3 + (x −b)3 + (x − c)3 − 3(x − a) (x − b) (x −c)

Recall the formula

`a^3 +b^3 +c^3 - 3abc = (a+b+c)(a^2 + b^2+ c^2  - ab - bc +_ ca)`

Using the above formula the given expression becomes

(x − a)3 + (x −b)3 + (x − c)3 − 3(x − a) (x − b) (x −c)

`{(x-a) +(x-b) + (x-c)} {(x-a)^2+(x-b)^2+(x-c)^2 -(x-a)(x-b)-(x-b)(x-c) - (x-c)(x-a)}`

`(x-a +x-b+x -c) {(x-a)^2+(x-b)^2+(x-c)^2 -(x-a)(x-b)-(x-b)(x-c) - (x-c)(x-a)} `

`(3x-a -b-c) {(x-a)^2+(x-b)^2+(x-c)^2 -(x-a)(x-b)-(x-b)(x-c) - (x-c)(x-a)}`

Given that

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

` ⇒ 3x -a-b-c =0`

Therefore the value of the given expression is

(x − a)3 + (x −b)3 + (x − c)3 − 3(x − a) (x − b) (x −c)

`= 0.{(x-a)^2+(x-b)^2+(x-c)^2 -(x-a)(x-b)-(x-b)(x-c) - (x-c)(x-a)}`

`=0`

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Chapter 5: Factorisation of Algebraic Expressions - Exercise 5.6 [Page 26]

APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 5 Factorisation of Algebraic Expressions
Exercise 5.6 | Q 13 | Page 26

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