simpify \( \frac{4^{y} \cdot 2^{y+2}-8^{y}}{2^{2} \cdot 3^{0} \cdot 2^{3 y}} \)
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Let's simplify the expression step-by-step! First, we notice that \(8^y\) can be rewritten in terms of base 2: \[ 8^y = (2^3)^y = 2^{3y}. \] Now, we rewrite the expression: \[ \frac{4^y \cdot 2^{y+2} - 2^{3y}}{2^2 \cdot 3^0 \cdot 2^{3y}}. \] Next, we can express \(4^y\) in terms of base 2: \[ 4^y = (2^2)^y = 2^{2y}. \] Now substituting for \(4^y\) we get: \[ \frac{2^{2y} \cdot 2^{y+2} - 2^{3y}}{2^2 \cdot 2^{3y}}. \] Combining the powers of 2 in the numerator: \[ 2^{2y} \cdot 2^{y+2} = 2^{2y+y+2} = 2^{3y+2}. \] So the numerator becomes: \[ 2^{3y+2} - 2^{3y}. \] Now, we can factor out \(2^{3y}\) from the numerator: \[ 2^{3y}(2^2 - 1) = 2^{3y}(4 - 1) = 2^{3y} \cdot 3. \] Now, we substitute it back into our expression: \[ \frac{2^{3y} \cdot 3}{2^2 \cdot 2^{3y}}. \] The \(2^{3y}\) terms will cancel out: \[ \frac{3}{2^2} = \frac{3}{4}. \] Thus, the simplified form of the given expression is: \[ \frac{3}{4}. \]
