\(\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)+15\)
\(\left(x^2+4x-5\right)\left(x^2+4x+3\right)+15\)
Đặt x2+4x+3=a, ta có:
\(\left(a-8\right)a+15\)= a2-8a+15=(a-4)2-1=(a-4+1)(a-4-1)=(a-3)(a-5)
Suy ra: (x2+4x+3-3)(x2+4x+3-5)=(x2+4x)(x2+4x-2)