If xy = 1, what is the value of 2^(x+y)^22^(x-y)^2?

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16 We apply exponent subtraction to combine the powers of two. 2^(x+y)^22^(x-y)^2 = 2^(x+y)^2 - (x-y)^2 We apply "Development Formulas - Squares of +/-" to expand each squared binomial. (x+y)^2 = x^2 + y^2 + 2xy (x-y)^2 = x^2 + y^2 - 2xy We subtract the expanded exponents to simplify the exponent expression. (x+y)^2 - (x-y)^2 = 4xy We substitute the given value into the exponent and evaluate the power of two. 2^4xy = 2^4 1 = 2^4 = 16

Explanation	Calculations	Help
$We apply exponent subtraction to combine the powers of two.$	$\frac{2 ^{(x + y)^{2}}}{2 ^{(x - y)^{2}}} = 2^{(x + y)^{2} - (x - y)^{2}}$
$We apply "Development Formulas - Squares of +/-" to expand each squared binomial.$	$(x + y)^{2} = x^{2} + y^{2} + 2 x y (x - y)^{2} = x^{2} + y^{2} - 2 x y$
$We subtract the expanded exponents to simplify the exponent expression.$	$(x + y)^{2} - (x - y)^{2} = 4 x y$
$We substitute the given value into the exponent and evaluate the power of two.$	$2^{4 x y} = 2^{4 \times 1} = 2^{4} = 16$

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