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4\times \frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}
Rationalize the denominator of \frac{1}{\sqrt{2}-1} by multiplying numerator and denominator by \sqrt{2}+1.
4\times \frac{\sqrt{2}+1}{\left(\sqrt{2}\right)^{2}-1^{2}}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}
Consider \left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
4\times \frac{\sqrt{2}+1}{2-1}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}
Square \sqrt{2}. Square 1.
4\times \frac{\sqrt{2}+1}{1}+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}
Subtract 1 from 2 to get 1.
4\left(\sqrt{2}+1\right)+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}
Anything divided by one gives itself.
4\sqrt{2}+4+\sqrt{3}\left(\sqrt{3}-\sqrt{6}\right)+\sqrt{8}
Use the distributive property to multiply 4 by \sqrt{2}+1.
4\sqrt{2}+4+\left(\sqrt{3}\right)^{2}-\sqrt{3}\sqrt{6}+\sqrt{8}
Use the distributive property to multiply \sqrt{3} by \sqrt{3}-\sqrt{6}.
4\sqrt{2}+4+3-\sqrt{3}\sqrt{6}+\sqrt{8}
The square of \sqrt{3} is 3.
4\sqrt{2}+4+3-\sqrt{3}\sqrt{3}\sqrt{2}+\sqrt{8}
Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.
4\sqrt{2}+4+3-3\sqrt{2}+\sqrt{8}
Multiply \sqrt{3} and \sqrt{3} to get 3.
4\sqrt{2}+7-3\sqrt{2}+\sqrt{8}
Add 4 and 3 to get 7.
\sqrt{2}+7+\sqrt{8}
Combine 4\sqrt{2} and -3\sqrt{2} to get \sqrt{2}.
\sqrt{2}+7+2\sqrt{2}
Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.
3\sqrt{2}+7
Combine \sqrt{2} and 2\sqrt{2} to get 3\sqrt{2}.