(a+1)的4次方+(a+1)的4次方-272=?

(a+1)^4+(a+3)^4-272
=(a+1)^4-256+(a+3)^4-16
=[ (a+1)^2-16 ][ (a+1)^2+16 ]+[ (a+3)^2-4 ][ (a+3)^2+4 ]
=(a+1+4)(a+1-4)(a^2+2a+17)+(a+3+2)(a+3-2)(a^2+6a+13)
=(a+5)[ (a-3)(a^2+2a+17)+(a+1)(a^2+6a+13) ]
=(a+5)(2a^3+6a^2+30a-38)
=2(a+5)(a^3+3a^2+15a-19)
=2(a+5)(a-1)(a^2+4a+19)

??问题打错了吧