A^3+b^3+3b^2+3b+1/a^2-ab-a+(b+1)^2 при а=4 - корень из 3, b=4+ корень из 3

$a\sqrt{3}+b\sqrt{3}+3b\sqrt{2}+3b+\frac{1}{a}\sqrt{2}-ab-a+(b+1)\sqrt{2}$

$\sqrt{3}(a+b)+\sqrt{2}(3b+\frac{1}{a}+b+1)+3b-ab-a$

1) \sqrt{3}(4-\sqrt{3}+4+\sqrt{3})=4\sqrt{3}\\2) \sqrt{2}(34+\sqrt{3}+\frac{1}{4-\sqrt{3}}+4+\sqrt{3}+1)=\sqrt{2}(17+4\sqrt{4}+\frac{1}{4-\sqrt{3}})=\\=17\sqrt{2}+4\sqrt{6}+\frac{\sqrt{2}}{4+\sqrt{3}}\\3) 12+3\sqrt{3}-16+3-4+\sqrt{3}=4\sqrt{3}-5\\4) $1) \sqrt{3}(4-\sqrt{3}+4+\sqrt{3})=4\sqrt{3}\\2) \sqrt{2}(3b+\frac{1}{4-\sqrt{3}}+4+\sqrt{3}+1)=\sqrt{2}(17+4\sqrt{4}+\frac{1}{4-\sqrt{3}})=\\=17\sqrt{2}+4\sqrt{6}+\frac{\sqrt{2}}{4-\sqrt{3}}\\3) 12+3\sqrt{3}-16+3-4+\sqrt{3}=4\sqrt{3}-5\\4) 4\sqrt{3}+4\sqrt{3}-5+17\sqrt{2}+4\sqrt{6}+\frac{\sqrt{2}}{4-\sqrt{3}}=\\=8\sqrt{3}+17\sqrt{2}+\frac{\sqrt{2}}{4-\sqrt{3}}=$ $\frac{32\sqrt{3}-24+68\sqrt{2}-17\sqrt{6}+\sqrt{2}}{4-\sqrt{3}}=\\=\frac{69\sqrt{2}+32\sqrt{3}-17\sqrt{6}-24}{4-\sqrt{3}}$

дальше я не знаю как делать...((