Solve (sqrt{2}-1)^0-(sqrt{5}-sqrt{3})^2*(sqrt{5}+sqrt{3})^2

$$( \sqrt { 2 } - 1 ) ^ { 0 } - ( \sqrt { 5 } - \sqrt { 3 } ) ^ { 2 } \times ( \sqrt { 5 } + \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$-3$$

Solution Steps

Calculate \sqrt{2}-1 to the power of 0 and get 1.

$$1-\left(\sqrt{5}-\sqrt{3}\right)^{2}\left(\sqrt{5}+\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-\sqrt{3}\right)^{2}.

$$1-\left(\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)\left(\sqrt{5}+\sqrt{3}\right)^{2}$$

The square of \sqrt{5} is 5.

$$1-\left(5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)\left(\sqrt{5}+\sqrt{3}\right)^{2}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$1-\left(5-2\sqrt{15}+\left(\sqrt{3}\right)^{2}\right)\left(\sqrt{5}+\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$1-\left(5-2\sqrt{15}+3\right)\left(\sqrt{5}+\sqrt{3}\right)^{2}$$

Add 5 and 3 to get 8.

$$1-\left(8-2\sqrt{15}\right)\left(\sqrt{5}+\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+\sqrt{3}\right)^{2}.

$$1-\left(8-2\sqrt{15}\right)\left(\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$1-\left(8-2\sqrt{15}\right)\left(5+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$1-\left(8-2\sqrt{15}\right)\left(5+2\sqrt{15}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$1-\left(8-2\sqrt{15}\right)\left(5+2\sqrt{15}+3\right)$$

Add 5 and 3 to get 8.

$$1-\left(8-2\sqrt{15}\right)\left(8+2\sqrt{15}\right)$$

Consider \left(8-2\sqrt{15}\right)\left(8+2\sqrt{15}\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 8.

$$1-\left(64-\left(2\sqrt{15}\right)^{2}\right)$$

Expand \left(2\sqrt{15}\right)^{2}.

$$1-\left(64-2^{2}\left(\sqrt{15}\right)^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$1-\left(64-4\left(\sqrt{15}\right)^{2}\right)$$

The square of \sqrt{15} is 15.

$$1-\left(64-4\times 15\right)$$

Multiply 4 and 15 to get 60.

$$1-\left(64-60\right)$$

Subtract 60 from 64 to get 4.

$$1-4$$

Subtract 4 from 1 to get -3.

$$-3$$

Factor

$$-3$$