1. **State the problem:** Solve for real $x$ in the equation $$\arctan\left(\frac{1}{n^2+n+1}\right) = \arctan\left(\frac{1}{n}\right) - \arctan(x).$$ 2. **Recall the formula for difference of arctan:** $$\arctan(a) - \arctan(b) = \arctan\left(\frac{a-b}{1+ab}\right)$$ provided the result is in the principal range. 3. **Apply the formula to the right side:** $$\arctan\left(\frac{1}{n}\right) - \arctan(x) = \arctan\left(\frac{\frac{1}{n} - x}{1 + \frac{1}{n}x}\right) = \arctan\left(\frac{\frac{1}{n} - x}{1 + \frac{x}{n}}\right).$$ 4. **Set the arguments equal since arctan is one-to-one on its principal domain:** $$\frac{1}{n^2+n+1} = \frac{\frac{1}{n} - x}{1 + \frac{x}{n}}.$$ 5. **Simplify the denominator:** $$1 + \frac{x}{n} = \frac{n+x}{n}.$$ 6. **Rewrite the right side:** $$\frac{\frac{1}{n} - x}{1 + \frac{x}{n}} = \frac{\frac{1}{n} - x}{\frac{n+x}{n}} = \left(\frac{1}{n} - x\right) \cdot \frac{n}{n+x} = \frac{1 - nx}{n+x}.$$ 7. **Equate and solve for $x$:** $$\frac{1}{n^2+n+1} = \frac{1 - nx}{n+x}.$$ Cross-multiplied: $$1 \cdot (n+x) = (1 - nx)(n^2+n+1).$$ 8. **Expand right side:** $$n + x = (1)(n^2+n+1) - nx(n^2+n+1) = n^2 + n + 1 - nx(n^2+n+1).$$ 9. **Bring all terms to one side:** $$n + x - n^2 - n - 1 + nx(n^2+n+1) = 0,$$ which simplifies to $$x + nx(n^2+n+1) - n^2 - 1 = 0.$$ 10. **Factor $x$ terms:** $$x(1 + n(n^2+n+1)) = n^2 + 1.$$ 11. **Simplify the coefficient of $x$:** $$1 + n(n^2+n+1) = 1 + n^3 + n^2 + n = n^3 + n^2 + n + 1.$$ 12. **Solve for $x$:** $$x = \frac{n^2 + 1}{n^3 + n^2 + n + 1}.$$ **Final answer:** $$\boxed{x = \frac{n^2 + 1}{n^3 + n^2 + n + 1}}.$$