Solve (3^{3}-2*9):3+{[(4^{3}-2*7-2^2*5):10]^2+1}:5+4^2-7^0

$$( 3 ^ { 3 } - 2 \times 9 ) : 3 + \{ [ ( 4 ^ { 3 } - 2 \times 7 - 2 ^ { 2 } \times 5 ) : 10 ] ^ { 2 } + 1 \} : 5 + 4 ^ { 2 } - 7 ^ { 0 }$$

Evaluate

$$20$$

Solution Steps

Calculate 3 to the power of 3 and get 27.

$$\frac{27-2\times 9}{3}+\frac{\left(\frac{4^{3}-2\times 7-2^{2}\times 5}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Multiply 2 and 9 to get 18.

$$\frac{27-18}{3}+\frac{\left(\frac{4^{3}-2\times 7-2^{2}\times 5}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Subtract 18 from 27 to get 9.

$$\frac{9}{3}+\frac{\left(\frac{4^{3}-2\times 7-2^{2}\times 5}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Divide 9 by 3 to get 3.

$$3+\frac{\left(\frac{4^{3}-2\times 7-2^{2}\times 5}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Calculate 4 to the power of 3 and get 64.

$$3+\frac{\left(\frac{64-2\times 7-2^{2}\times 5}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Multiply 2 and 7 to get 14.

$$3+\frac{\left(\frac{64-14-2^{2}\times 5}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Subtract 14 from 64 to get 50.

$$3+\frac{\left(\frac{50-2^{2}\times 5}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Calculate 2 to the power of 2 and get 4.

$$3+\frac{\left(\frac{50-4\times 5}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Multiply 4 and 5 to get 20.

$$3+\frac{\left(\frac{50-20}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Subtract 20 from 50 to get 30.

$$3+\frac{\left(\frac{30}{10}\right)^{2}+1}{5}+4^{2}-7^{0}$$

Divide 30 by 10 to get 3.

$$3+\frac{3^{2}+1}{5}+4^{2}-7^{0}$$

Calculate 3 to the power of 2 and get 9.

$$3+\frac{9+1}{5}+4^{2}-7^{0}$$

Add 9 and 1 to get 10.

$$3+\frac{10}{5}+4^{2}-7^{0}$$

Divide 10 by 5 to get 2.

$$3+2+4^{2}-7^{0}$$

Add 3 and 2 to get 5.

$$5+4^{2}-7^{0}$$

Calculate 4 to the power of 2 and get 16.

$$5+16-7^{0}$$

Add 5 and 16 to get 21.

$$21-7^{0}$$

Calculate 7 to the power of 0 and get 1.

$$21-1$$

Subtract 1 from 21 to get 20.

$$20$$

Factor

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