1. **State the problem:** Simplify the expression $\left(x+\frac{1}{3}\right)\left(x-\frac{1}{3}\right) - \left(-\frac{2}{3} - 2x\right)^2$. 2. **Recall formulas:** - The product of conjugates: $\left(a+b\right)\left(a-b\right) = a^2 - b^2$. - Square of a binomial: $\left(a+b\right)^2 = a^2 + 2ab + b^2$. 3. **Apply the product of conjugates formula:** $$\left(x+\frac{1}{3}\right)\left(x-\frac{1}{3}\right) = x^2 - \left(\frac{1}{3}\right)^2 = x^2 - \frac{1}{9}$$ 4. **Expand the square:** $$\left(-\frac{2}{3} - 2x\right)^2 = \left(-2x - \frac{2}{3}\right)^2 = \left(-2x\right)^2 + 2\cdot(-2x)\cdot\left(-\frac{2}{3}\right) + \left(-\frac{2}{3}\right)^2$$ Calculate each term: - $\left(-2x\right)^2 = 4x^2$ - $2\cdot(-2x)\cdot\left(-\frac{2}{3}\right) = 2 \cdot (-2x) \cdot (-\frac{2}{3}) = \frac{8}{3}x$ - $\left(-\frac{2}{3}\right)^2 = \frac{4}{9}$ So, $$\left(-\frac{2}{3} - 2x\right)^2 = 4x^2 + \frac{8}{3}x + \frac{4}{9}$$ 5. **Substitute back into the original expression:** $$x^2 - \frac{1}{9} - \left(4x^2 + \frac{8}{3}x + \frac{4}{9}\right)$$ 6. **Distribute the minus sign:** $$x^2 - \frac{1}{9} - 4x^2 - \frac{8}{3}x - \frac{4}{9}$$ 7. **Combine like terms:** - Combine $x^2$ terms: $x^2 - 4x^2 = -3x^2$ - Combine constants: $-\frac{1}{9} - \frac{4}{9} = -\frac{5}{9}$ So the expression becomes: $$-3x^2 - \frac{8}{3}x - \frac{5}{9}$$ **Final answer:** $$\boxed{-3x^2 - \frac{8}{3}x - \frac{5}{9}}$$