Solve (2^2*2^3)-sqrt[3]{8*27}+3^-2=

$$( 2 ^ { 2 } \cdot 2 ^ { 3 } ) - \sqrt[ 3 ] { 8 \cdot 27 } + 3 ^ { - 2 } =$$

Evaluate

$$\frac{235}{9}\approx 26.111111111$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 3 to get 5.

$$2^{5}-\sqrt[3]{8\times 27}+3^{-2}$$

Calculate 2 to the power of 5 and get 32.

$$32-\sqrt[3]{8\times 27}+3^{-2}$$

Multiply 8 and 27 to get 216.

$$32-\sqrt[3]{216}+3^{-2}$$

Calculate \sqrt[3]{216} and get 6.

$$32-6+3^{-2}$$

Subtract 6 from 32 to get 26.

$$26+3^{-2}$$

Calculate 3 to the power of -2 and get \frac{1}{9}.

$$26+\frac{1}{9}$$

Add 26 and \frac{1}{9} to get \frac{235}{9}.

$$\frac{235}{9}$$

Factor

$$\frac{5 \cdot 47}{3 ^ {2}} = 26\frac{1}{9} = 26.11111111111111$$