Solve 3(6-(9+(1+3)(1+3)div2-20))

$$3(6-(9+(1+3)(1+3) \div 2-20))$$

Evaluate

$$27$$

Solution Steps

Multiply 1+3 and 1+3 to get \left(1+3\right)^{2}.

$$3\left(6-\left(9+\frac{\left(1+3\right)^{2}}{2}-20\right)\right)$$

Add 1 and 3 to get 4.

$$3\left(6-\left(9+\frac{4^{2}}{2}-20\right)\right)$$

Calculate 4 to the power of 2 and get 16.

$$3\left(6-\left(9+\frac{16}{2}-20\right)\right)$$

Divide 16 by 2 to get 8.

$$3\left(6-\left(9+8-20\right)\right)$$

Add 9 and 8 to get 17.

$$3\left(6-\left(17-20\right)\right)$$

Subtract 20 from 17 to get -3.

$$3\left(6-\left(-3\right)\right)$$

The opposite of -3 is 3.

$$3\left(6+3\right)$$

Add 6 and 3 to get 9.

$$3\times 9$$

Multiply 3 and 9 to get 27.

$$27$$

Factor

$$3^{3}$$