Solve (2bc)^-3/3^2b^2c^2=

$$\frac { ( 2 b c ) ^ { - 3 } } { 3 ^ { 2 } b ^ { 2 } c ^ { 2 } } =$$

Evaluate

$$\frac{1}{72\left(bc\right)^{5}}$$

Solution Steps

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

$$\frac{2^{-3}b^{-3}c^{-3}}{3^{2}b^{2}c^{2}}$$

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

$$\frac{\frac{1}{8}b^{-3}c^{-3}}{3^{2}b^{2}c^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\frac{1}{8}b^{-3}c^{-3}}{9b^{2}c^{2}}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{\frac{1}{8}}{9b^{5}c^{5}}$$

Express \frac{\frac{1}{8}}{9b^{5}c^{5}} as a single fraction.

$$\frac{1}{8\times 9b^{5}c^{5}}$$

Multiply 8 and 9 to get 72.

$$\frac{1}{72b^{5}c^{5}}$$

Expand

$$\frac{1}{72\left(bc\right)^{5}}$$

Solution Steps

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

$$\frac{2^{-3}b^{-3}c^{-3}}{3^{2}b^{2}c^{2}}$$

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

$$\frac{\frac{1}{8}b^{-3}c^{-3}}{3^{2}b^{2}c^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\frac{1}{8}b^{-3}c^{-3}}{9b^{2}c^{2}}$$

To divide powers of the same base, subtract the numerator's exponent from the denominator's exponent.

$$\frac{\frac{1}{8}}{9b^{5}c^{5}}$$

Express \frac{\frac{1}{8}}{9b^{5}c^{5}} as a single fraction.

$$\frac{1}{8\times 9b^{5}c^{5}}$$

Multiply 8 and 9 to get 72.

$$\frac{1}{72b^{5}c^{5}}$$