Ta có
a3+b3+c3-3abc
=(a+b)3-3ab(a+b)+c3-3abc
=[(a+b)3+c3]-3ab(a+b+c)
=(a+b+c)[(a+b)2-c(a+b)+c2]-3ab(a+b+c)
=(a=b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)
=(a+b+c)(a2+2ab+b2-ac-bc+c2-3ab)
=(a+b+c)(a2+b2+c2-ab-ac-bc)
a3+b3+c3-3abc
=(a+b)3-3ab(a+b)+c3-3abc
=[(a+b)3+c3]-3ab(a+b+c)
=(a+b+c)[(a+b)2-c(a+b)+c2]-3ab(a+b+c)
=(a=b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)
=(a+b+c)(a2+2ab+b2-ac-bc+c2-3ab)
=(a+b+c)(a2+b2+c2-ab-ac-bc)
a3 + b3 + c3 - 3abc
= [( a3 + b3 ) + c3 ] - 3abc
= [( a + b )3 + c3 + 3ab( a + b )] - 3abc
= [( a + b )3 + c3 ] + 3a2b + 3ab2 - 3abc
= ( a + b + c ) [( a + b )2 - c( a + b ) + c2 ] - 3ab( a + b + c )
= ( a + b + c ) ( a2 + 2ab + b2 - ac - bc + c2 - 3ab )
= ( a + b + c ) ( a2 + b2 + c2 - ab - bc - ac )
= 1/2( a + b + c ) ( 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac )
= 1/2( a + b + c ) [( a2 - 2ab + b2 ) + ( a2 - 2ac + c2 ) + ( b2 - 2bc + c2 )]
= 1/2 ( a + b + c ) [( a - b )2 + ( a - c )2 + ( b - c )2]
Hok Tốt!!!
