Solve (2^4-3*4)+sqrt[3]{8}*3+sqrt{5^2-3^2}

$$( { 2 }^{ 4 } -3 \times 4)+ \sqrt[3]{ 8 } \times 3+ \sqrt{ { 5 }^{ 2 } - { 3 }^{ 2 } }$$

Evaluate

$$14$$

Solution Steps

Calculate 2 to the power of 4 and get 16.

$$16-3\times 4+\sqrt[3]{8}\times 3+\sqrt{5^{2}-3^{2}}$$

Multiply 3 and 4 to get 12.

$$16-12+\sqrt[3]{8}\times 3+\sqrt{5^{2}-3^{2}}$$

Subtract 12 from 16 to get 4.

$$4+\sqrt[3]{8}\times 3+\sqrt{5^{2}-3^{2}}$$

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

$$4+2\times 3+\sqrt{5^{2}-3^{2}}$$

Multiply 2 and 3 to get 6.

$$4+6+\sqrt{5^{2}-3^{2}}$$

Add 4 and 6 to get 10.

$$10+\sqrt{5^{2}-3^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$10+\sqrt{25-3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$10+\sqrt{25-9}$$

Subtract 9 from 25 to get 16.

$$10+\sqrt{16}$$

Calculate the square root of 16 and get 4.

$$10+4$$

Add 10 and 4 to get 14.

$$14$$

Factor

$$2\times 7$$