Solve {(2/3)^6*(2/3)^-3}+(4/9)^2

$$\{ ( \frac { 2 } { 3 } ) ^ { 6 } \times ( \frac { 2 } { 3 } ) ^ { - 3 } \} + ( \frac { 4 } { 9 } ) ^ { 2 }$$

Evaluate

$$\frac{40}{81}\approx 0.49382716$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 6 and -3 to get 3.

$$\left(\frac{2}{3}\right)^{3}+\left(\frac{4}{9}\right)^{2}$$

Calculate \frac{2}{3} to the power of 3 and get \frac{8}{27}.

$$\frac{8}{27}+\left(\frac{4}{9}\right)^{2}$$

Calculate \frac{4}{9} to the power of 2 and get \frac{16}{81}.

$$\frac{8}{27}+\frac{16}{81}$$

Add \frac{8}{27} and \frac{16}{81} to get \frac{40}{81}.

$$\frac{40}{81}$$

Factor

$$\frac{2 ^ {3} \cdot 5}{3 ^ {4}} = 0.49382716049382713$$