Solve (6sqrt{2}-frac{4sqrt{2}}{x})^2

$$( 6 \sqrt { 2 } - \frac { 4 \sqrt { 2 } } { x } ) ^ { 2 }$$

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 6\sqrt{2} times \frac{x}{x}.

$$\left(\frac{6\sqrt{2}x}{x}-\frac{4\sqrt{2}}{x}\right)^{2}$$

Since \frac{6\sqrt{2}x}{x} and \frac{4\sqrt{2}}{x} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{6\sqrt{2}x-4\sqrt{2}}{x}\right)^{2}$$

To raise \frac{6\sqrt{2}x-4\sqrt{2}}{x} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(6\sqrt{2}x-4\sqrt{2}\right)^{2}}{x^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6\sqrt{2}x-4\sqrt{2}\right)^{2}.

$$\frac{36\left(\sqrt{2}\right)^{2}x^{2}-48\sqrt{2}x\sqrt{2}+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{36\left(\sqrt{2}\right)^{2}x^{2}-48\times 2x+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{36\times 2x^{2}-48\times 2x+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

Multiply 36 and 2 to get 72.

$$\frac{72x^{2}-48\times 2x+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

Multiply -48 and 2 to get -96.

$$\frac{72x^{2}-96x+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{72x^{2}-96x+16\times 2}{x^{2}}$$

Multiply 16 and 2 to get 32.

$$\frac{72x^{2}-96x+32}{x^{2}}$$

Expand

$$72-\frac{96}{x}+\frac{32}{x^{2}}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 6\sqrt{2} times \frac{x}{x}.

$$\left(\frac{6\sqrt{2}x}{x}-\frac{4\sqrt{2}}{x}\right)^{2}$$

Since \frac{6\sqrt{2}x}{x} and \frac{4\sqrt{2}}{x} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{6\sqrt{2}x-4\sqrt{2}}{x}\right)^{2}$$

To raise \frac{6\sqrt{2}x-4\sqrt{2}}{x} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(6\sqrt{2}x-4\sqrt{2}\right)^{2}}{x^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6\sqrt{2}x-4\sqrt{2}\right)^{2}.

$$\frac{36\left(\sqrt{2}\right)^{2}x^{2}-48\sqrt{2}x\sqrt{2}+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{36\left(\sqrt{2}\right)^{2}x^{2}-48\times 2x+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{36\times 2x^{2}-48\times 2x+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

Multiply 36 and 2 to get 72.

$$\frac{72x^{2}-48\times 2x+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

Multiply -48 and 2 to get -96.

$$\frac{72x^{2}-96x+16\left(\sqrt{2}\right)^{2}}{x^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{72x^{2}-96x+16\times 2}{x^{2}}$$

Multiply 16 and 2 to get 32.

$$\frac{72x^{2}-96x+32}{x^{2}}$$