Solve 3^2*(1/27)^2div(1/9)^3=

$$3 ^ { 2 } \times ( \frac { 1 } { 27 } ) ^ { 2 } \div ( \frac { 1 } { 9 } ) ^ { 3 } =$$

Evaluate

$$9$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$\frac{9\times \left(\frac{1}{27}\right)^{2}}{\left(\frac{1}{9}\right)^{3}}$$

Calculate \frac{1}{27} to the power of 2 and get \frac{1}{729}.

$$\frac{9\times \frac{1}{729}}{\left(\frac{1}{9}\right)^{3}}$$

Multiply 9 and \frac{1}{729} to get \frac{9}{729}.

$$\frac{\frac{9}{729}}{\left(\frac{1}{9}\right)^{3}}$$

Reduce the fraction \frac{9}{729} to lowest terms by extracting and canceling out 9.

$$\frac{\frac{1}{81}}{\left(\frac{1}{9}\right)^{3}}$$

Calculate \frac{1}{9} to the power of 3 and get \frac{1}{729}.

$$\frac{\frac{1}{81}}{\frac{1}{729}}$$

Divide \frac{1}{81} by \frac{1}{729} by multiplying \frac{1}{81} by the reciprocal of \frac{1}{729}.

$$\frac{1}{81}\times 729$$

Multiply \frac{1}{81} and 729 to get \frac{729}{81}.

$$\frac{729}{81}$$

Divide 729 by 81 to get 9.

$$9$$

Factor

$$3^{2}$$