Question 1018·Hard·Math

Type $y=(x+6)^3-(x-4)^3$ and $y=1000$. Click both intersections: Desmos shows $(-6,1000)$ and $(4,1000)$. Each intersection’s $x$-coordinate solves the equation. So the real solutions are $-6$ and $4$; $1000$ is the value both sides share, not a solution.

The two $x$-values are distinct, meaning different, so include each once. Add them:

The sum of the distinct real solutions is $-2$. Grid in -2.

The inputs $x+6$ and $x-4$ are $10$ apart. Their midpoint, or number halfway between them, is:

So $x+6=u+5$ and $x-4=u-5$. The equal shifts make cancellation easier to see.

Use the cube expansion $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ on each input:

Keep the minus sign in front of the second cube when you subtract next.

Subtract every term in the second expansion:

The cubic and linear terms cancel, so the given equation becomes $30u^2+250=1000$.

Subtract $250$ from both sides:

Divide both sides by $30$:

Take both square roots because $5^2$ and $(-5)^2$ both equal $25$:

Since $u=x+1$, subtract $1$ from each value of $u$:

Both values are different real solutions, so add each once:

The sum of the distinct real solutions is $-2$. Grid in -2.