Odpowiedź:
[tex]a)\ \ (5x-1)^3=(5x)^3-3\cdot(5x)^2\cdot1+3\cdot5x\cdot1^2-1^3=125x^3-3\cdot25x^2\cdot1+3\cdot5x\cdot1-1=\\\\=125x^3-75x^2+15x-1\\\\\\b)\ \ (3x+\frac{1}{2}y)^3=(3x)^3+3\cdot(3x)^2\cdot\frac{1}{2}y+3\cdot3x\cdot(\frac{1}{2}y)^2+(\frac{1}{2}y)^3=\\\\=27x^3+3\cdot9x^2\cdot\frac{1}{2}y+9x\cdot\frac{1}{4}y^2+\frac{1}{8}y^3=27x^3+27x^2\cdot\frac{1}{2}y+\frac{9}{4}xy^2+\frac{1}{8}y^3=\\\\=27x^3+\frac{27}{2}x^2y+\frac{9}{4}xy^2+\frac{1}{8}y^3[/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]
