Question

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

Answer 1

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`