After you write $(x+6)^3 - (x-4)^3$ as $(a-b)(a^2 + ab + b^2)$ with $a = x+6$ and $b = x-4$, simplify the $(a-b)$ part and divide both sides of the equation by that factor.

When you expand $(x+6)^2$, $(x+6)(x-4)$, and $(x-4)^2$, combine like terms to get a quadratic equation. Then solve that quadratic.

Once you find the two real solutions of the quadratic, do not stop there. Make sure to add those two solution values together, since the question asks for their sum.

In Desmos, enter y = (x + 6)^3 - (x - 4)^3 on one line and y = 1000 on another line so you can see where the two graphs intersect.

Zoom or pan as needed until you see both intersection points between the two graphs. Tap each intersection point and note its x-coordinate; these are the solutions to the equation.

Take the two x-values you noted from the intersection points and add them together (you can type an expression like x1 + x2 in Desmos using those values, or just add them manually) to get the requested sum.

Recognize that the left side is a difference of cubes.

Use the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

Let $a = x+6$ and $b = x-4$. Then

Simplify the first factor:

so the equation becomes

Now divide both sides by 10 to simplify.

After dividing both sides by 10, you get

Now expand each term:

• $(x+6)^2 = x^2 + 12x + 36$
• $(x+6)(x-4) = x^2 + 2x - 24$
• $(x-4)^2 = x^2 - 8x + 16$

Add them together:

So the equation becomes

Subtract 100 from both sides:

Divide the whole equation by 3 to simplify:

Now you have a quadratic equation whose solutions are the same as the original equation’s solutions.

Factor the quadratic:

You need two numbers that multiply to $-24$ and add to $2$. Those numbers are $6$ and $-4$, so

Thus the solutions are $x = -6$ and $x = 4$.

The question asks for the sum of the distinct real solutions, so compute

Therefore, the sum of the distinct real solutions is $-2$.