[tex]3x^{3} -3=7x^{2} -7x\\\\3(x^{3} -1)=7x(x -1)\\\\3(x^{3} -1)-7x(x -1)=0\\\\3(x^{3} -1^{3} )-7x(x -1)=0\\\\korzystam ~~ze~~wzoru~~skroconego~~mnozenia:\\x^{3} -y^{3} =(x-y)\cdot (x^{2} +xy+y^{2} )\\\\3(x-1)\cdot (x^{2} +x+1)-7x(x-1)=0\\\\(x-1)\cdot [3\cdot (x^{2} +x+1)-7x]=0\\\\(x-1)\cdot (3x^{2} +3x+3-7x)=0\\\\(x-1)\cdot (3x^{2} -4x+3)=0\\\\3x^{2} -4x+3=0~~\lor~~x-1=0\\\\3x^{2} -4x+3=0~~\lor~~x=1\\\\[/tex]
[tex]a=3,~~b=-4,~~c=3\\\Delta=b^{2} -4\cdot a \cdot c\\\Delta=(-4)^{2} -4\cdot 3\cdot 3\\\Delta=16-36\\\Delta=-20~~\Rightarrow~~\Delta < 0 ~~\Rightarrow~~brak~~pierwiastkow~~\Rightarrow~~x\in \oslash\\\\\\[/tex]
Odp: Rozwiązaniem równaia jest x = 1.
