Odpowiedź:
[tex]a)\\\\(x+1)^3-(x-1)^3=6x^2-x\\\\x^3+3x^2\cdot1+3x\cdot1^2+1^3-(x^3-3x^2\cdot1+3x\cdot1^2-1^3)=6x^2-x\\\\x^3+3x^2+3x+1-(x^3-3x^2+3x-1)=6x^2-x\\\\x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2-x\\\\6x^2+2=6x^2-x\\\\6x^2-6x^2+x=-2\\\\x=-2[/tex]
[tex]b)\\\\(x+2)^3-(x+1)^3=3x^2+x\\\\x^3+3x^2\cdot2+3x\cdot2^2+2^3-(x^3+3x^2\cdot1+3x\cdot1^2+1^3)=3x^2+x\\\\x^3+6x^2+3x \cdot4+8-(x^3+3x^2+3x+1)=3x^2+x\\\\x^3+6x^2+12x+8-x^3-3x^2-3x-1=3x^2+x\\\\3x^2+9x+7=3x^2+x\\\\3x^2+9x-3x^2-x=-7\\\\8x=-7\ \ /:8\\\\x=-\frac{7}{8}[/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]
