Răspuns:
(n+1)(n+2)(n+3)+1 = (n²+3n+1)²
(n²+2n+n+2)×(n+3)+1 = n⁴+9n²+1+6n³+2n²+6n
(n²+3n+2)×(n+3)+1 = n⁴+9n²+1+6n³+2n²+6n
n³+3n²+3n²+9n+2n+6+1 =n⁴+9n²+1+6n³+2n²+6n
n³+3n²+3n²+9n+2n+6 =n⁴+9n²+6n³+2n²+6n
n³+6n²+11n+6 = n⁴+11n²+6n³+6n
n³+6n²+11n+6-n⁴-11n²-6n³-6n =0
-5n³-5n²+5n+6-n⁴=0
-n⁴-5n³-5n²+5n+6=0
-n⁴+n³-6n³+6n²-11n²+11n-6n+6 = 0
-n³(n-1)-6n²(n-1)-11n(n-1)-6(n-1) = 0
-(n-1)(n³+6n²+11n+6) = 0
-(n-1)(n³+n²+5n²+5n+6n+6) = 0
-(n-1)×(n²×(n+1)+5n×(n+1)+6×(n+1)) = 0
-(n-1)×(n+1)×(n²+5n+6) = 0
-(n-1)×(n+1)×(n²+3n+2n+6) = 0
-(n-1)×(n+1)×(n×(n+3)+2×(n+3)) = 0
-(n-1)×(n+1)×(n+3)×(n+2)= 0
(n-1)×(n+1)×(n+3)×(n+2)= 0
n-1=0 => n=1
n+1=0 => n=-1
n+3=0 => n=-3
n+2=0 => n=-2
Sper ca te-am ajutat :)
