Solve (2x^-3y^2)^3*2x^0

$$( 2 x ^ { - 3 } y ^ { 2 } ) ^ { 3 } \cdot 2 x ^ { 0 }$$

Evaluate

$$\frac{16y^{6}}{x^{9}}$$

Solution Steps

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

$$2^{3}\left(x^{-3}\right)^{3}\left(y^{2}\right)^{3}\times 2x^{0}$$

To raise a power to another power, multiply the exponents. Multiply -3 and 3 to get -9.

$$2^{3}x^{-9}\left(y^{2}\right)^{3}\times 2x^{0}$$

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

$$2^{3}x^{-9}y^{6}\times 2x^{0}$$

Calculate 2 to the power of 3 and get 8.

$$8x^{-9}y^{6}\times 2x^{0}$$

Multiply 8 and 2 to get 16.

$$16x^{-9}y^{6}x^{0}$$

Calculate x to the power of 0 and get 1.

$$16x^{-9}y^{6}\times 1$$

Multiply 16 and 1 to get 16.

$$16x^{-9}y^{6}$$

Differentiate w.r.t. x

$$-\frac{144y^{6}}{x^{10}}$$