6.025. $(x-a)^{3}-(x-b)^{3}=b^{3}-a^{3}$.
## Solution.

Domain of definition: $x \in R$.

Let's factorize the left and right parts of the equation as differences of cubes:

$$
(x-a-x+b)\left((x-a)^{2}+(x-a)(x-b)+(x-b)^{2}\right)=(b-a)\left(b^{2}+b a+a^{2}\right)
$$

$(b-a)\left(x^{2}-2 a x+a^{2}+x^{2}-(a+b) x+a b+x^{2}-2 b x+b^{2}\right)-$ $-(b-a)\left(b^{2}+b a+a^{2}\right)=0$

$(b-a)\left(3 x^{2}-3(a+b) x+a b+a^{2}+b^{2}\right)-(b-a)\left(b^{2}+b a+a^{2}\right)=0 \Leftrightarrow$ $\Leftrightarrow(b-a)\left(x^{2}-(a+b) x\right)=0$.

## From this:

1) if $b-a=0, b=a$, then $x \in R$;
2) if $b-a \neq 0, b \neq a$, then $x^{2}-(a+b) x=0$, or $x(x-(a+b))=0$, from which $x_{1}=0, x_{2}=a+b$.

Answer: if $a=b$, then $x \in R$; if $a \neq b$, then $x_{1}=0, x_{2}=a+b$.