A simple observation would do!

Let $(x-3) = a$ and $(x-7) = b$ , then $(2x-10) = a + b$

We have the equation $a^3 + b^3 = (a+b)^3$

Exapnding the RHS, $a^3 + b^3 = a^3 + b^3 + 3ab(a+b)$

$3ab(a+b) = 0$

Solutions are $a= 0, b=0$ and $a + b = 0$

Therefore $x = 3, x = 7$ and $x = 5$

The sum of real roots is $3+ 5 + 7 = \boxed{15}$

Let $y = x - 5$ . Then the equation becomes

$(y + 2)^{3} + (y - 2)^{3} = (2y)^{3} = 8y^{3}$ .

Now when we expand the two terms on the LHS we have several terms cancel, leaving us with

$\Longrightarrow 2y^{3} + 2*\binom{3}{2}*4y = 8y^{3} \Longrightarrow 6y(y^{2} - 4) = 0$ ,

which has solutions $y = 0, 2, -2$ , which sum to $0$ .

But since $x = y + 5$ , the real roots of the original equation will be $3, 5$ and $7$ , which sum to $\boxed{15}$ .

Alternatively, going back to the equation $(y + 2)^{3} + (y - 2)^{3} = 8y^{3}$ , we can observe that if $y = a$ is a solution then $y = -a$ is also a solution, since

$(-a + 2)^{3} + (-a - 2)^{3} = -(a - 2)^{3} - (a + 2)^{3} = -((a + 2)^{3} + (a - 2)^{3}) = -8a^{3} = 8*(-a)^{3}$ .

This, coupled with the observation that $y = 0$ is a solution, implies that the sum of the roots to this equation is $a + (-a) + 0 = 0$ , and thus the sum of the roots to the original equation is $3*5 = \boxed{15}$ .

Let $f(x) = (x-3)^3 + (x-7)^3 - (2x-10)^3$ .

Then we note that:

$f(3) = 0 + (-4)^3 - (-4)^3 = 0 +64-64 = 0$

$f(5) = 2^3 + (-2)^3 - 0 = 8-8-0 = 0$

$f(7) = 4^3+0-4^3 = 64+0-64 = 0$

Since $f(x)$ is a polynomial with a degree of $3$ , it can only have maximum $3$ roots. Therefore, the roots are $3$ , $5$ and $7$ and the sum of roots $= 3+5+7 = \boxed{15}$

Let $(x-3)=a$ , $(x-7)=b$ and $(10-2x)=c$ . Frist we have $a+b+c=0$ , so $a^{3}+b^{3}+c^{3}=3abc$ . Therefore $abc=0 => x_{1}=3; x_{2}=7; x_{3}=5$

Let(x-3)=a,(x-7)=b then (2x-10)=a+b or -c. So,transferring 2x-10 to L.H.S. we get, a^3 +b^3+(-a-b)^3=0.Here,a+b+c=a+b-a-b=0. So,a^3+b^3+c^3=3 a.b.c, as a+b+c=0 Therefore 0=3.(x-3)(x-7).(2x-10) So, x=3,7,5.Hence,sum of real roots is 15.

Sum of roots=-b/a To find b and a find coefficient of x^2 and x^3 respectively. So value of a=6 and value of b=-90 hence sum of roots=- (-90)/6=15