Odpowiedź:
[tex]a)\ \ (1+\sqrt{2})^3+(1-\sqrt{2})^3=\\\\=1^3+3*1^2*\sqrt{2}+3*1*(\sqrt{2})^2+(\sqrt{2})^3+(1^3-3*1^2*\sqrt{2}+3*1*(\sqrt{2})^2-(\sqrt{2})^3=\\\\=1+3*1*\sqrt{2}+3*2+\sqrt{8}+(1-3*1*\sqrt{2}+3*2-\sqrt{8})=\\\\=1+3\sqrt{2}+6+\sqrt{8}+(1-3\sqrt{2}+6-\sqrt{8})=7+3\sqrt{2}+\sqrt{8}+1-3\sqrt{2}+6-\sqrt{8}=14[/tex]
[tex]b)\ \ (4+\sqrt{2})^3-(4-\sqrt{2})^3=\\\\=4^3+3*4^2*\sqrt{2}+3*4*(\sqrt{2})^2+(\sqrt{2})^3-(4^3-3*4^2*\sqrt{2}+3*4*(\sqrt{2})^2-(\sqrt{2})^3=\\\\=64+3*16\sqrt{2}+12*2+\sqrt{8}-(64-3*16\sqrt{2}+12*2-\sqrt{8})=\\\\=64+48\sqrt{2}+24+\sqrt{8}-(64-48\sqrt{2}+24-\sqrt{8})=\\\\=88+48\sqrt{2}+\sqrt{8}-64+48\sqrt{2}-24+\sqrt{8}=88+96\sqrt{2}+2\sqrt{8}-88=\\\\=96\sqrt{2}+2\sqrt{4*2}=96\sqrt{2}+2*2\sqrt{2}=96\sqrt{2}+4\sqrt{2}=100\sqrt{2}[/tex]
[tex]Zastosowane\ \ wzory\\\\(a+b)^3=a^3+3a^2b+3ab^2+b^3\\\\a-b)=a^3-3a^2b+3ab^2-b^3[/tex]
