Solve [12div4+(6)(3-1)]^2+sqrt[3]{27}+sqrt[3]{8}=

$$[ 12 \div 4 + ( 6 ) ( 3 - 1 ) ] ^ { 2 } + \sqrt[ 3 ] { 27 } + \sqrt[ 3 ] { 8 } =$$

Evaluate

$$230$$

Solution Steps

Divide 12 by 4 to get 3.

$$\left(3+6\left(3-1\right)\right)^{2}+\sqrt[3]{27}+\sqrt[3]{8}$$

Subtract 1 from 3 to get 2.

$$\left(3+6\times 2\right)^{2}+\sqrt[3]{27}+\sqrt[3]{8}$$

Multiply 6 and 2 to get 12.

$$\left(3+12\right)^{2}+\sqrt[3]{27}+\sqrt[3]{8}$$

Add 3 and 12 to get 15.

$$15^{2}+\sqrt[3]{27}+\sqrt[3]{8}$$

Calculate 15 to the power of 2 and get 225.

$$225+\sqrt[3]{27}+\sqrt[3]{8}$$

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

$$225+3+\sqrt[3]{8}$$

Add 225 and 3 to get 228.

$$228+\sqrt[3]{8}$$

Calculate \sqrt[3]{8} and get 2.

$$228+2$$

Add 228 and 2 to get 230.

$$230$$

Factor

$$2\times 5\times 23$$