1. **State the problem:** Solve the equation $$(x-1)^2 + (-x-1)^2 = 0$$ 2. **Recall the formula and rules:** The square of any real number is always non-negative, i.e., $a^2 \geq 0$ for all real $a$. The sum of two squares equals zero only if each square is zero individually. 3. **Apply this to the problem:** $$ (x-1)^2 + (-x-1)^2 = 0 \implies (x-1)^2 = 0 \text{ and } (-x-1)^2 = 0 $$ 4. **Solve each equation:** - From $(x-1)^2 = 0$, we get $x-1=0 \implies x=1$ - From $(-x-1)^2 = 0$, we get $-x-1=0 \implies -x=1 \implies x=-1$ 5. **Check if both conditions can be true simultaneously:** Since the sum of squares is zero only if both are zero, $x$ must satisfy both $x=1$ and $x=-1$ simultaneously, which is impossible. 6. **Conclusion:** No real value of $x$ satisfies the equation because the two conditions contradict each other. **Final answer:** No real solution.