1. **State the problem:** Simplify the expression $$(x^2 - xy - 2y^2)^2$$. 2. **Recall the formula:** The square of a binomial $(a - b)^2 = a^2 - 2ab + b^2$. Here, the expression inside the square is a trinomial, so we first simplify inside the parentheses if possible. 3. **Factor the inner expression:** Factor $x^2 - xy - 2y^2$. We look for two numbers that multiply to $-2$ and add to $-1$ (coefficient of $-xy$): those numbers are $-2$ and $1$. So, $$x^2 - xy - 2y^2 = (x - 2y)(x + y)$$. 4. **Rewrite the original expression:** $$((x - 2y)(x + y))^2 = (x - 2y)^2 (x + y)^2$$. 5. **Apply the square to each factor:** $$ (x - 2y)^2 = x^2 - 2 \cdot x \cdot 2y + (2y)^2 = x^2 - 4xy + 4y^2 $$ $$ (x + y)^2 = x^2 + 2xy + y^2 $$ 6. **Multiply the two quadratics:** $$ (x^2 - 4xy + 4y^2)(x^2 + 2xy + y^2) $$ 7. **Expand step-by-step:** - $x^2 \cdot x^2 = x^4$ - $x^2 \cdot 2xy = 2x^3y$ - $x^2 \cdot y^2 = x^2y^2$ - $-4xy \cdot x^2 = -4x^3y$ - $-4xy \cdot 2xy = -8x^2y^2$ - $-4xy \cdot y^2 = -4xy^3$ - $4y^2 \cdot x^2 = 4x^2y^2$ - $4y^2 \cdot 2xy = 8xy^3$ - $4y^2 \cdot y^2 = 4y^4$ 8. **Combine like terms:** - $x^4$ - $2x^3y - 4x^3y = -2x^3y$ - $x^2y^2 - 8x^2y^2 + 4x^2y^2 = (-3x^2y^2)$ - $-4xy^3 + 8xy^3 = 4xy^3$ - $4y^4$ 9. **Final simplified expression:** $$x^4 - 2x^3y - 3x^2y^2 + 4xy^3 + 4y^4$$