Find the Domain (x^2+2xy+y^2)*(x+y)=64

Problem

(x^2+2*x*y+y^2)*(x+y)=64

Solution

1. Recognize the perfect square trinomial within the first set of parentheses.

(x+y)^2=x^2+2*x*y+y^2

2. Substitute this identity back into the original equation.

(x+y)^2*(x+y)=64

3. Simplify the expression by combining the powers of the common base (x+y)

(x+y)^3=64

4. Solve for the sum x+y by taking the cube root of both sides.

√(3,(x+y)^3)=√(3,64)

x+y=4

5. Identify the domain of the relation. Since the equation simplifies to a linear equation y=−x+4 there are no restrictions such as division by zero or square roots of negative numbers. The variable x can take any real value.

Final Answer

Domain: *(−∞,∞)

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