Solve (frac{8x^{2}}{y})^-1(x^3/4y^-1)^-2

$$( \frac { 8 x ^ { 2 } } { y } ) ^ { - 1 } ( \frac { x ^ { 3 } } { 4 y ^ { - 1 } } ) ^ { - 2 }$$

Evaluate

$$\frac{2}{yx^{8}}$$

Solution Steps

To raise \frac{8x^{2}}{y} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(8x^{2}\right)^{-1}}{y^{-1}}\times \left(\frac{x^{3}}{4y^{-1}}\right)^{-2}$$

To raise \frac{x^{3}}{4y^{-1}} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(8x^{2}\right)^{-1}}{y^{-1}}\times \frac{\left(x^{3}\right)^{-2}}{\left(4y^{-1}\right)^{-2}}$$

Multiply \frac{\left(8x^{2}\right)^{-1}}{y^{-1}} times \frac{\left(x^{3}\right)^{-2}}{\left(4y^{-1}\right)^{-2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(8x^{2}\right)^{-1}\left(x^{3}\right)^{-2}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

To raise a power to another power, multiply the exponents. Multiply 3 and -2 to get -6.

$$\frac{\left(8x^{2}\right)^{-1}x^{-6}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

Expand \left(8x^{2}\right)^{-1}.

$$\frac{8^{-1}\left(x^{2}\right)^{-1}x^{-6}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

To raise a power to another power, multiply the exponents. Multiply 2 and -1 to get -2.

$$\frac{8^{-1}x^{-2}x^{-6}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

Calculate 8 to the power of -1 and get \frac{1}{8}.

$$\frac{\frac{1}{8}x^{-2}x^{-6}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

To multiply powers of the same base, add their exponents. Add -2 and -6 to get -8.

$$\frac{\frac{1}{8}x^{-8}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

Expand \left(4y^{-1}\right)^{-2}.

$$\frac{\frac{1}{8}x^{-8}}{y^{-1}\times 4^{-2}\left(y^{-1}\right)^{-2}}$$

To raise a power to another power, multiply the exponents. Multiply -1 and -2 to get 2.

$$\frac{\frac{1}{8}x^{-8}}{y^{-1}\times 4^{-2}y^{2}}$$

Calculate 4 to the power of -2 and get \frac{1}{16}.

$$\frac{\frac{1}{8}x^{-8}}{y^{-1}\times \frac{1}{16}y^{2}}$$

To multiply powers of the same base, add their exponents. Add -1 and 2 to get 1.

$$\frac{\frac{1}{8}x^{-8}}{y^{1}\times \frac{1}{16}}$$

Calculate y to the power of 1 and get y.

$$\frac{\frac{1}{8}x^{-8}}{y\times \frac{1}{16}}$$

Expand

$$\frac{2}{yx^{8}}$$

Solution Steps

To raise \frac{8x^{2}}{y} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(8x^{2}\right)^{-1}}{y^{-1}}\times \left(\frac{x^{3}}{4y^{-1}}\right)^{-2}$$

To raise \frac{x^{3}}{4y^{-1}} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(8x^{2}\right)^{-1}}{y^{-1}}\times \frac{\left(x^{3}\right)^{-2}}{\left(4y^{-1}\right)^{-2}}$$

Multiply \frac{\left(8x^{2}\right)^{-1}}{y^{-1}} times \frac{\left(x^{3}\right)^{-2}}{\left(4y^{-1}\right)^{-2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(8x^{2}\right)^{-1}\left(x^{3}\right)^{-2}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

To raise a power to another power, multiply the exponents. Multiply 3 and -2 to get -6.

$$\frac{\left(8x^{2}\right)^{-1}x^{-6}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

Expand \left(8x^{2}\right)^{-1}.

$$\frac{8^{-1}\left(x^{2}\right)^{-1}x^{-6}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

To raise a power to another power, multiply the exponents. Multiply 2 and -1 to get -2.

$$\frac{8^{-1}x^{-2}x^{-6}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

Calculate 8 to the power of -1 and get \frac{1}{8}.

$$\frac{\frac{1}{8}x^{-2}x^{-6}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

To multiply powers of the same base, add their exponents. Add -2 and -6 to get -8.

$$\frac{\frac{1}{8}x^{-8}}{y^{-1}\times \left(4y^{-1}\right)^{-2}}$$

Expand \left(4y^{-1}\right)^{-2}.

$$\frac{\frac{1}{8}x^{-8}}{y^{-1}\times 4^{-2}\left(y^{-1}\right)^{-2}}$$

To raise a power to another power, multiply the exponents. Multiply -1 and -2 to get 2.

$$\frac{\frac{1}{8}x^{-8}}{y^{-1}\times 4^{-2}y^{2}}$$

Calculate 4 to the power of -2 and get \frac{1}{16}.

$$\frac{\frac{1}{8}x^{-8}}{y^{-1}\times \frac{1}{16}y^{2}}$$

To multiply powers of the same base, add their exponents. Add -1 and 2 to get 1.

$$\frac{\frac{1}{8}x^{-8}}{y^{1}\times \frac{1}{16}}$$

Calculate y to the power of 1 and get y.

$$\frac{\frac{1}{8}x^{-8}}{y\times \frac{1}{16}}$$