Question

因式分解..a的四次方加b的四次加c的四次减2a方b方减2b方c方减2倍c方a方..？

Answer 1

a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2
=(a^4-2a^2b^2+b^4)-2a^2c^2+2b^2c^2+c^4-4b^2c^2
=(a^2-b^2)^2-2c^2*(a^2-b^2)+c^4-4b^2c^2
=(a^2-b^2-c^2)^2-4b^2c^2
=(a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)
=[a^2-(b^2-2bc+c^2)][a^2-(b^2＋2bc＋c^2)]
=[a^2-(b－c)^2][a^2-(b＋c)^2]
=(a+b-c)(a-b+c)(a+b+c)(a-b-c),4,In[3]:= Factor[a^4+b^4+c^4-2*a^2*b^2-2*b^2*c^2-2*c^2*a^2]
Out[3]= (a - b - c) (a + b - c) (a - b + c) (a + b + c)
用Mathematica5算得...
具体过程嘛，
a^4+b^4+c^4-2*a^2*b^2-2*b^2*c^2-2*c^2*a^2,2,a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2
=(a^4-2a^2b^2+b^4)-2a^2c^2+2b^2c^2+c^4-4b^2c^2
=(a^2-b^2)^2-2c^2*(a^2-b^2)+c^4-4b^2c^2
=(a^2-b^2-c^2)^2-4b^2c^2
=(a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc...,0,a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2
=(a^2-b^2-c^2)^2-4b^2c^2
=(a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)
=[a^2-（b^2-2bc+c^2）][a^2-（b^2＋2bc＋c^2)]
=[a^2-（b－c）^2][a^2-（b＋c)^2]
=(a+b-c)(a-b+c)(a+b+c)(a-b-c),0,