Question

Find the value of (x+a)^3+(x-b)^3+(x-c)^3-(x-a)(x-b)(x-c) where a+b+c=3x

Answer 1

Answer:

Answer = 0

Step-by-step explanation:

We know that,

a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)

If a+b+c=0 then,

a³+b³+c³-3abc=0×(a²+b²+c²-ab-bc-ca)=0 .....(1)

(x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)

let a=(x-a)

b=(x-b)

and

c=(x-c)

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

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

Thus the answer is 0.