If the expression 2x^2y^-(1)/(4)sqrt [4](16x^2y^4) is written in the form ax^by^c then what is the product of a, b and c? square

To simplify the expression $2x^{2}y^{-\frac {1}{4}}\sqrt [4]{16x^{2}y^{4}}$, we need to rewrite the expression in the form $ax^{b}y^{c}$.<br /><br />First, let's simplify the expression inside the fourth root:<br />$\sqrt [4]{16x^{2}y^{4}} = \sqrt [4]{16} \cdot \sqrt [4]{x^{2}} \cdot \sqrt [4]{y^{4}}$<br /><br />We know that $\sqrt [4]{16} = 2$, $\sqrt [4]{x^{2}} = x^{\frac{1}{2}}$, and $\sqrt [4]{y^{4}} = y$. So, we can rewrite the expression as:<br />$\sqrt [4]{16x^{2}y^{4}} = 2 \cdot x^{\frac{1}{2}} \cdot y$<br /><br />Now, let's substitute this back into the original expression:<br />$2x^{2}y^{-\frac {1}{4}} \cdot 2 \cdot x^{\frac{1}{2}} \cdot y$<br /><br />Next, we can combine the coefficients and the variables separately:<br />$2 \cdot 2 \cdot x^{2} \cdot x^{\frac{1}{2}} \cdot y^{-\frac {1}{4}} \cdot y$<br /><br />Simplifying the coefficients, we get:<br />$4 \cdot x^{2 + \frac{1}{2}} \cdot y^{-\frac {1}{4}} \cdot y$<br /><br />Combining the exponents of $x$ and $y$, we get:<br />$4 \cdot x^{\frac{5}{2}} \cdot y^{\frac{3}{4}}$<br /><br />Therefore, the expression $2x^{2}y^{-\frac {1}{4}}\sqrt [4]{16x^{2}y^{4}}$ can be written in the form $ax^{b}y^{c}$, where $a = 4$, $b = \frac{5}{2}$, and $c = \frac{3}{4}$.<br /><br />The product of $a$, $b$, and $c$ is:<br />$4 \cdot \frac{5}{2} \cdot \frac{3}{4} = 15/2$<br /><br />So, the product of $a$, $b$, and $c$ is $\boxed{15/2}$.