Solve (sqrt{7}-sqrt{5})^2(sqrt{7}+sqrt{5})^2

$$( \sqrt { 7 } - \sqrt { 5 } ) ^ { 2 } ( \sqrt { 7 } + \sqrt { 5 } ) ^ { 2 }$$

Evaluate

$$4$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{7}-\sqrt{5}\right)^{2}.

$$\left(\left(\sqrt{7}\right)^{2}-2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}\right)\left(\sqrt{7}+\sqrt{5}\right)^{2}$$

The square of \sqrt{7} is 7.

$$\left(7-2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}\right)\left(\sqrt{7}+\sqrt{5}\right)^{2}$$

To multiply \sqrt{7} and \sqrt{5}, multiply the numbers under the square root.

$$\left(7-2\sqrt{35}+\left(\sqrt{5}\right)^{2}\right)\left(\sqrt{7}+\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$\left(7-2\sqrt{35}+5\right)\left(\sqrt{7}+\sqrt{5}\right)^{2}$$

Add 7 and 5 to get 12.

$$\left(12-2\sqrt{35}\right)\left(\sqrt{7}+\sqrt{5}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7}+\sqrt{5}\right)^{2}.

$$\left(12-2\sqrt{35}\right)\left(\left(\sqrt{7}\right)^{2}+2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}\right)$$

The square of \sqrt{7} is 7.

$$\left(12-2\sqrt{35}\right)\left(7+2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}\right)$$

To multiply \sqrt{7} and \sqrt{5}, multiply the numbers under the square root.

$$\left(12-2\sqrt{35}\right)\left(7+2\sqrt{35}+\left(\sqrt{5}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$\left(12-2\sqrt{35}\right)\left(7+2\sqrt{35}+5\right)$$

Add 7 and 5 to get 12.

$$\left(12-2\sqrt{35}\right)\left(12+2\sqrt{35}\right)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 12.

$$144-\left(2\sqrt{35}\right)^{2}$$

Expand \left(2\sqrt{35}\right)^{2}.

$$144-2^{2}\left(\sqrt{35}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$144-4\left(\sqrt{35}\right)^{2}$$

The square of \sqrt{35} is 35.

$$144-4\times 35$$

Multiply 4 and 35 to get 140.

$$144-140$$

Subtract 140 from 144 to get 4.

$$4$$

Factor

$$2^{2}$$