Solve (2x^3y^2)^2div(-1/2xy)^3

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

Evaluate

$$-32yx^{3}$$

Solution Steps

Expand \left(2x^{3}y^{2}\right)^{2}.

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

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

$$\frac{2^{2}x^{6}\left(y^{2}\right)^{2}}{\left(-\frac{1}{2}xy\right)^{3}}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{2^{2}x^{6}y^{4}}{\left(-\frac{1}{2}xy\right)^{3}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4x^{6}y^{4}}{\left(-\frac{1}{2}xy\right)^{3}}$$

Expand \left(-\frac{1}{2}xy\right)^{3}.

$$\frac{4x^{6}y^{4}}{\left(-\frac{1}{2}\right)^{3}x^{3}y^{3}}$$

Calculate -\frac{1}{2} to the power of 3 and get -\frac{1}{8}.

$$\frac{4x^{6}y^{4}}{-\frac{1}{8}x^{3}y^{3}}$$

Cancel out x^{3}y^{3} in both numerator and denominator.

$$\frac{4yx^{3}}{-\frac{1}{8}}$$

Divide 4yx^{3} by -\frac{1}{8} by multiplying 4yx^{3} by the reciprocal of -\frac{1}{8}.

$$\frac{4yx^{3}\times 8}{-1}$$

Multiply 4 and 8 to get 32.

$$\frac{32yx^{3}}{-1}$$

Anything divided by -1 gives its opposite.

$$-32yx^{3}$$

Expand

$$-32yx^{3}$$

Solution Steps

Expand \left(2x^{3}y^{2}\right)^{2}.

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

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

$$\frac{2^{2}x^{6}\left(y^{2}\right)^{2}}{\left(-\frac{1}{2}xy\right)^{3}}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{2^{2}x^{6}y^{4}}{\left(-\frac{1}{2}xy\right)^{3}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{4x^{6}y^{4}}{\left(-\frac{1}{2}xy\right)^{3}}$$

Expand \left(-\frac{1}{2}xy\right)^{3}.

$$\frac{4x^{6}y^{4}}{\left(-\frac{1}{2}\right)^{3}x^{3}y^{3}}$$

Calculate -\frac{1}{2} to the power of 3 and get -\frac{1}{8}.

$$\frac{4x^{6}y^{4}}{-\frac{1}{8}x^{3}y^{3}}$$

Cancel out x^{3}y^{3} in both numerator and denominator.

$$\frac{4yx^{3}}{-\frac{1}{8}}$$

Divide 4yx^{3} by -\frac{1}{8} by multiplying 4yx^{3} by the reciprocal of -\frac{1}{8}.

$$\frac{4yx^{3}\times 8}{-1}$$

Multiply 4 and 8 to get 32.

$$\frac{32yx^{3}}{-1}$$

Anything divided by -1 gives its opposite.

$$-32yx^{3}$$