因式分解(a^2+b^2+c^2+d^2-ab-ac-ad-bc-bd-cd)(a+b+c+d)^2指平方

Question

因式分解(a^2+b^2+c^2+d^2-ab-ac-ad-bc-bd-cd)(a+b+c+d)
^2指平方

Answer 1

( a" + b" + c" + d" - ab - ac - ad - bc - bd - cd )( a + b + c + d )
= ( a+ b+ c +d )[ ( a" + 2ab + b" ) - 3ab - 3cd + ( c" + 2cd + d" ) - ac - ad - bc - bd ]
= ( a+ b+ c +d )[ ( a + b )" - 3ab - 3cd + ( c + d )" - a( c + d ) - b( c + d ) ]
= [ ( a + b ) + ( c + d ) ][ ( a + b )" - ( a + b )( c + d ) + ( c + d )" - 3ab - 3cd ]

= ( a + b )"' + ( c + d )"' - 3ab( a + b ) - 3cd( a + b ) - 3ab( c + d ) - 3cd( c + d )
= ( a + b )"' + ( c"' + 3c"d + 3cd" + d" ) - 3ab( a + b ) - 3cd( a + b ) - 3ab( c + d ) - 3c"d - 3cd"
= ( a"' + 3a"b + 3ab" + b"' ) + c"' + d"' - 3a"b - 3ab" - 3ab( c + d ) - 3cd( a + b )
= a"' + b"' + c"' + d"' - 3abc - 3abd - 3acd - 3bcd

= a"' + b"' - 3cd( a + b ) + c"' + d"' - 3ab( c + d )
= ( a + b )( a" - ab + b" ) - 3cd( a + b ) + ( c + d )( c" - cd + d" ) - 3ab( c + d )
= ( a + b )( a" - ab + b" - 3cd ) + ( c + d )( c" - cd + d" - 3ab )