1. **State the problem:** Find the derivative of the function $$y = (1 - x)^4 (1 + x + x^2)^4$$. 2. **Formula used:** We will use the product rule for differentiation: $$\frac{d}{dx}[u \cdot v] = u'v + uv'$$ where $$u = (1 - x)^4$$ and $$v = (1 + x + x^2)^4$$. 3. **Differentiate each part:** - For $$u = (1 - x)^4$$, use the chain rule: $$u' = 4(1 - x)^3 \cdot \frac{d}{dx}(1 - x) = 4(1 - x)^3 (-1) = -4(1 - x)^3$$. - For $$v = (1 + x + x^2)^4$$, again use the chain rule: $$v' = 4(1 + x + x^2)^3 \cdot \frac{d}{dx}(1 + x + x^2) = 4(1 + x + x^2)^3 (1 + 2x)$$. 4. **Apply the product rule:** $$\frac{dy}{dx} = u'v + uv' = -4(1 - x)^3 (1 + x + x^2)^4 + (1 - x)^4 \cdot 4(1 + x + x^2)^3 (1 + 2x)$$. 5. **Factor common terms:** Both terms have $$4(1 - x)^3 (1 + x + x^2)^3$$, so factor it out: $$\frac{dy}{dx} = 4(1 - x)^3 (1 + x + x^2)^3 \left[-(1 + x + x^2) + (1 - x)(1 + 2x)\right]$$. 6. **Simplify inside the brackets:** Calculate each part: - $$-(1 + x + x^2) = -1 - x - x^2$$ - $$(1 - x)(1 + 2x) = 1 + 2x - x - 2x^2 = 1 + x - 2x^2$$ Sum: $$-1 - x - x^2 + 1 + x - 2x^2 = (-1 + 1) + (-x + x) + (-x^2 - 2x^2) = 0 + 0 - 3x^2 = -3x^2$$ 7. **Final derivative:** $$\frac{dy}{dx} = 4(1 - x)^3 (1 + x + x^2)^3 (-3x^2) = -12x^2 (1 - x)^3 (1 + x + x^2)^3$$. **Answer:** $$\boxed{\frac{dy}{dx} = -12x^2 (1 - x)^3 (1 + x + x^2)^3}$$