Solve (frac{3^{2}+6-16}{1})^2+9(5-4)^3|+11^2-(-6+3)^2

$$( \frac { 3 ^ { 2 } + 6 - 16 } { 1 } ) ^ { 2 } + 9 ( 5 - 4 ) ^ { 3 } | + 11 ^ { 2 } - ( - 6 + 3 ) ^ { 2 }$$

Evaluate

$$1009$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$\left(\frac{9+6-16}{1}\right)^{2}+9\left(5-4\right)^{3}|11^{2}-\left(-6+3\right)^{2}|$$

Add 9 and 6 to get 15.

$$\left(\frac{15-16}{1}\right)^{2}+9\left(5-4\right)^{3}|11^{2}-\left(-6+3\right)^{2}|$$

Subtract 16 from 15 to get -1.

$$\left(\frac{-1}{1}\right)^{2}+9\left(5-4\right)^{3}|11^{2}-\left(-6+3\right)^{2}|$$

Anything divided by one gives itself.

$$\left(-1\right)^{2}+9\left(5-4\right)^{3}|11^{2}-\left(-6+3\right)^{2}|$$

Calculate -1 to the power of 2 and get 1.

$$1+9\left(5-4\right)^{3}|11^{2}-\left(-6+3\right)^{2}|$$

Subtract 4 from 5 to get 1.

$$1+9\times 1^{3}|11^{2}-\left(-6+3\right)^{2}|$$

Calculate 1 to the power of 3 and get 1.

$$1+9\times 1|11^{2}-\left(-6+3\right)^{2}|$$

Multiply 9 and 1 to get 9.

$$1+9|11^{2}-\left(-6+3\right)^{2}|$$

Calculate 11 to the power of 2 and get 121.

$$1+9|121-\left(-6+3\right)^{2}|$$

Add -6 and 3 to get -3.

$$1+9|121-\left(-3\right)^{2}|$$

Calculate -3 to the power of 2 and get 9.

$$1+9|121-9|$$

Subtract 9 from 121 to get 112.

$$1+9|112|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 112 is 112.

$$1+9\times 112$$

Multiply 9 and 112 to get 1008.

$$1+1008$$

Add 1 and 1008 to get 1009.

$$1009$$

Factor

$$1009$$