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

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

Evaluate

$$14$$

Solution Steps

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(\sqrt{2}+1\right)^{3}.

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

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

$$\left(\sqrt{2}\right)^{3}+6+3\sqrt{2}+1-\left(\sqrt{2}-1\right)^{3}$$

Add 6 and 1 to get 7.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\sqrt{2}-1\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(\sqrt{2}-1\right)^{3}.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\left(\sqrt{2}\right)^{3}-3\left(\sqrt{2}\right)^{2}+3\sqrt{2}-1\right)$$

The square of \sqrt{2} is 2.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\left(\sqrt{2}\right)^{3}-3\times 2+3\sqrt{2}-1\right)$$

Multiply -3 and 2 to get -6.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\left(\sqrt{2}\right)^{3}-6+3\sqrt{2}-1\right)$$

Subtract 1 from -6 to get -7.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\left(\sqrt{2}\right)^{3}-7+3\sqrt{2}\right)$$

To find the opposite of \left(\sqrt{2}\right)^{3}-7+3\sqrt{2}, find the opposite of each term.

$$\left(\sqrt{2}\right)^{3}+7+3\sqrt{2}-\left(\sqrt{2}\right)^{3}+7-3\sqrt{2}$$

Combine \left(\sqrt{2}\right)^{3} and -\left(\sqrt{2}\right)^{3} to get 0.

$$7+3\sqrt{2}+7-3\sqrt{2}$$

Add 7 and 7 to get 14.

$$14+3\sqrt{2}-3\sqrt{2}$$

Combine 3\sqrt{2} and -3\sqrt{2} to get 0.

$$14$$

Factor

$$2\times 7$$