Odpowiedź:
[tex]a)\\\\(2a+1)^3=(2a)^3+3\cdot(2a)^2\cdot1+3\cdot2a\cdot1^2+1^3=8a^3+3\cdot4a^2\cdot1+6a\cdot1+1=\\\\=8a^3+12a^2+6a+1\\\\\\b)\\\\(5-3x)^3=5^3-3\cdot5^2\cdot3x+3\cdot5\cdot(3x)^2-(3x)^3=125-3\cdot25\cdot3x+15\cdot9x^2-27x^3=\\\\=125-225x+135x^2-27x^3[/tex]
[tex]c)\\\\(x-\sqrt{2})^3=x^3-3x^2\sqrt{2}+3x\cdot(\sqrt{2})^2-(\sqrt{2})^3=x^3-3\sqrt{2}x^2+3x\cdot2-\sqrt{8}=\\\\=x^3-3\sqrt{2}x^2+6x-\sqrt{4\cdot2}=x^3-3\sqrt{2}x^2+6x-2\sqrt{2}\\\\\\d)\\\\(3x+\sqrt{5})^3=(3x)^3+3\cdot(3x)^2\cdot\sqrt{5}+3\cdot3x\cdot(\sqrt{5})^2+(\sqrt{5})^3=\\\\=27x^3+3\cdot9x^2\sqrt{5}+9x\cdot5+\sqrt{125}=27x^3+27\sqrt{5}x^2+45x+\sqrt{25\cdot5}=\\\\=27x^3+27\sqrt{5}x^2+45x+5\sqrt{5}[/tex]
[tex]Zastosowane\ \ wzory\\\\(a+b)^3=a^3+3a^2b+3ab^2+b^3\\\\(a-b)^3=a^3-3a^2b+3ab^2-b^3[/tex]
