1. Let's state the problem: Verify the given algebraic identity: $$((sx+y)(z^2+w^2))^2 = (s+y)^2 + 2(s+y)(z^2+w^2) + (z^2+w^2)^2$$ 2. First, let's expand the left side: $$((sx + y)(z^2 + w^2))^2 = ((sx + y)(z^2 + w^2))\cdot((sx + y)(z^2 + w^2))$$ 3. Multiply inside the parentheses: $$(sx + y)(z^2 + w^2) = sx \cdot z^2 + sx \cdot w^2 + y \cdot z^2 + y \cdot w^2 = sx z^2 + sx w^2 + y z^2 + y w^2$$ 4. So, the expression becomes: $$(sx z^2 + sx w^2 + y z^2 + y w^2)^2$$ 5. This is a square of a sum of four terms; expanding will be lengthy and includes cross terms. 6. Now examine the right side: $$(s + y)^2 + 2(s + y)(z^2 + w^2) + (z^2 + w^2)^2$$ 7. Notice that the right side resembles the expansion of $(a + b)^2 = a^2 + 2ab + b^2$ where $a = s + y$ and $b = z^2 + w^2$. 8. But the left side has $(sx + y)$ instead of $(s + y)$, so the expressions are not the same unless $x=1$. 9. Therefore the original equality holds only if $x=1$ and $s$ replaces $sx$ accordingly. 10. Hence the correct identity for general $x$ is: $$((s x + y)(z^2 + w^2))^2 \neq (s + y)^2 + 2(s + y)(z^2 + w^2) + (z^2 + w^2)^2$$ unless $x=1$ and the expression is: $$( (s + y)(z^2 + w^2))^2 = (s + y)^2 (z^2 + w^2)^2 = (s + y)^2 (z^2 + w^2)^2$$ which is not equal to the right side either, because the right side is a sum of squares and cross terms rather than a product squared. 11. Therefore, the given equality is false in general. Final answer: The equation $$((s x + y)(z^2 + w^2))^2 = (s + y)^2 + 2(s + y)(z^2 + w^2) + (z^2 + w^2)^2$$ is false for general variables.