Question

prove that:a³+b³+c³-3abc=½(a+b+c){(a-b)²+(b-c)²+(c-a)²}

Answer 1

a³+b³+c³-3abc=1\2(a+b+c){(a-b)²+(b-c)²+(c-a)²}
LHS
1\2(a+b+c){(a²-2ab+b²+b²-2bc+c²+c²-2ca+a²)}
1\2(a+b+c){(2a²+2b²+2c²)(-2ab-2bc-2ac)}
1\2×2(a+b+c)(a²+b²+c²)(-ab-bc-ac)
(a+b+c)(a²+b²+c²)(-ab-bc-ac)
a(a²+b²+c²)(-ab-bc-ac)+b(a²+b²+c²)(-ab-bc-ac)+c(a²+b²+c²)(-ab-bc-ac)
a³+ab²+ac²-ab-abc-ac+a²b+b³+bc²-ab-bc-abc+a²c+b²c+c³-abc-bc-ac
a³+b³+c³-3abc
all -ve ,+ve no. cut together u can solve in copy by seeing my answer
h.p