Find the value of (a−2x)
3
+(b−2x)
3
+(c−2x)
3
−3(a−2x)(b−2x)(c−2x), given that a+b+c=6x.

We are given the relation:
a+b+c=6x
We can rewrite this expression by moving 6x to the left side:

(a−2x)+(b−2x)+(c−2x)=a+b+c−6x
Substituting a+b+c=6x into the equation:

(a−2x)+(b−2x)+(c−2x)=6x−6x=0
Let p=a−2x, q=b−2x, and r=c−2x. We have established that:

p+q+r=0
Using the standard algebraic identity, if p+q+r=0, then:

p
3
+q
3
+r
3
−3pqr=0
Substituting back the values of p, q, and r:

(a−2x)
3
+(b−2x)
3
+(c−2x)
3
−3(a−2x)(b−2x)(c−2x)=0