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प्रश्न
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
विकल्प
a + b + c
(a − b) (b − c) (c − a)
0
none of these
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उत्तर
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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