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

$$( \sqrt[ 3 ] { 2 ^ { 2 } \cdot 3 \cdot 2 + 3 } - 2 ^ { 3 } : \sqrt[ 3 ] { - 8 } ) ^ { 2 }$$

Evaluate

$$49$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

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

Calculate 2 to the power of 3 and get 8.

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

Multiply 8 and 3 to get 24.

$$\left(\sqrt[3]{24+3}-\frac{2^{3}}{\sqrt[3]{-8}}\right)^{2}$$

Add 24 and 3 to get 27.

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

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

$$\left(3-\frac{2^{3}}{\sqrt[3]{-8}}\right)^{2}$$

Calculate 2 to the power of 3 and get 8.

$$\left(3-\frac{8}{\sqrt[3]{-8}}\right)^{2}$$

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

$$\left(3-\frac{8}{-2}\right)^{2}$$

Divide 8 by -2 to get -4.

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

The opposite of -4 is 4.

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

Add 3 and 4 to get 7.

$$7^{2}$$

Calculate 7 to the power of 2 and get 49.

$$49$$

Factor

$$7^{2}$$