((a3+4a2+10a+12):(a3-a2+2a+16))((a3-3a2+8a):(a2+2a+6))

[tex] \frac{ a^{3}+4a^2+10a+12}{a^3-a^2+2a+16} * \frac{a^3-3a^2+8a}{a^2+2a+6} = \frac{(a + 2) (a^2 + 2 a + 6)}{(a + 2) (a^2 - 3 a + 8)} * \frac{a(a^2-3a+8)}{a^2+2a+6} = a[/tex]


a^{3}+4a^2+10a+12 = a^3+2a^2+6a+2a^2+4a+12=a(a^2+2a+6)+2(a^2+2a+6)

a^3-a^2+2a+16 = a^3-3a^2+8a+2a^2-6a+16=a(a^2-3a+8)+2(a^2-3a+8)=(a+2)(a^2-3a+8)