Question

Descompunand in factori (x-3)³-x-3 obtinem?

Answer 1

 

$\displaystyle\\(x-3)^3-x-3=\\\\=x^3-3\cdot x^2\cdot3+3x\cdot(-3)^2-3^3-x-3=\\\\=x^3-9x^2+27x-27-x-3=\\\\=x^3-9x^2+26x-30=~~~\text{Acum il descompunem.}\\\\=x^3-5x^2-4x^2+20x+6x-30=~~~\text{Dam factor comun.}\\\\=x^2(x-5)-4x(x-5)+6(x-5)=\\\\=\boxed{(x-5)(x^2-4x+6)}$

 

 

 

Answer 2

$\it (x-3)^3-x-3= (x-3)^3-8+8-x-3= (x-3)^3-2^3-x+5=\\ \\ =(x-5)[(x-3)^2+2(x-3)+4]-(x-5)=(x-5)(x^2-6x+9+2x-6+4-1)=\\ \\ =(x-5)(x^2-4x+6)$