Solve [sqrt[3]{-27}-(-5+2)^0](4^2-3^1+7^3)

$$[ \sqrt[ 3 ] { - 27 } - ( - 5 + 2 ) ^ { 0 } ] ( 4 ^ { 2 } - 3 ^ { 1 } + 7 ^ { 3 } )$$

Evaluate

$$-1424$$

Solution Steps

Calculate \sqrt[3]{-27} and get -3.

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

Add -5 and 2 to get -3.

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

Calculate -3 to the power of 0 and get 1.

$$\left(-3-1\right)\left(4^{2}-3^{1}+7^{3}\right)$$

Subtract 1 from -3 to get -4.

$$-4\left(4^{2}-3^{1}+7^{3}\right)$$

Calculate 4 to the power of 2 and get 16.

$$-4\left(16-3^{1}+7^{3}\right)$$

Calculate 3 to the power of 1 and get 3.

$$-4\left(16-3+7^{3}\right)$$

Subtract 3 from 16 to get 13.

$$-4\left(13+7^{3}\right)$$

Calculate 7 to the power of 3 and get 343.

$$-4\left(13+343\right)$$

Add 13 and 343 to get 356.

$$-4\times 356$$

Multiply -4 and 356 to get -1424.

$$-1424$$

Factor

$$-1424$$