Question

descompuneti in factori:
1 supra lui 8 t^3+(1+1supra lui 2t)^3
1-(z+1)^6

Answer 1

t³/8+(1+t/2)³=
=(t/2)³+(1+t/2)³=                      formula este a³+b³=(a+b)(a²-ab+b²)
=(t/2+1+t/2)[(t/2)²-(t/2)(1+t/2)+(1+t/2)²]=
=(t+1)(t²/4-t/2-t²/4+1+2t/2+t²/4)= 
=(t+1)(t²/4+t/2+1) 

1-(z+1)⁶=[1- (z+1)³][1+ (z+1)³]=             prima formula folosita a²-b²=(a-b)(a+b)
={(1-z-1)[(1+z+1+(z+1)²]}{(1+z+1)[(1-z-1+(z+1)²]}=
=[(-z)(2+z+z²+2z+1)][(2+z)(-z+z²+2z+1)]=
=(-z)(z²+3z+3)(2+z)(z²+z+1)