Solve 3+{2^{2}+3*3^{2}-[-2^3:2^2]}+(3^3)^2=

$$3 + \{ 2 ^ { 2 } + 3 \cdot 3 ^ { 2 } - [ - 2 ^ { 3 } : 2 ^ { 2 } ] \} + ( 3 ^ { 3 } ) ^ { 2 } =$$

Evaluate

$$765$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

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

$$3+2^{2}+3^{3}-\frac{-2^{3}}{2^{2}}+3^{6}$$

Calculate 2 to the power of 2 and get 4.

$$3+4+3^{3}-\frac{-2^{3}}{2^{2}}+3^{6}$$

Add 3 and 4 to get 7.

$$7+3^{3}-\frac{-2^{3}}{2^{2}}+3^{6}$$

Calculate 3 to the power of 3 and get 27.

$$7+27-\frac{-2^{3}}{2^{2}}+3^{6}$$

Add 7 and 27 to get 34.

$$34-\frac{-2^{3}}{2^{2}}+3^{6}$$

Calculate 2 to the power of 3 and get 8.

$$34-\frac{-8}{2^{2}}+3^{6}$$

Calculate 2 to the power of 2 and get 4.

$$34-\frac{-8}{4}+3^{6}$$

Divide -8 by 4 to get -2.

$$34-\left(-2\right)+3^{6}$$

The opposite of -2 is 2.

$$34+2+3^{6}$$

Add 34 and 2 to get 36.

$$36+3^{6}$$

Calculate 3 to the power of 6 and get 729.

$$36+729$$

Add 36 and 729 to get 765.

$$765$$

Factor

$$3^{2}\times 5\times 17$$