Please answer 19 i) and ii)

Please answer 19 i) and ii) — — I yqo + P) + (91 ' KJ"duns ut •qj 10 q »eo s.'ojse,l JO tuns pull s.»qn» m.' Jumnut xuo•ut •BOI ve

$a^{3} (b - c)^{3} + b^{3} (c - a)^{3} + c^{3} (a - b)^{3} = {[a (b - c)]}^{3} + [b (c - a)]^{3} + [c (a - b)]^{3}$

We know that if a + b + c =0 then $a^{3} + b^{3} + c^{3} = 3 a b c$

$\Rightarrow a + b + c = [a (b - c)] + b (c - a) + c (a - b) = a b - a c + b c - a b + a c - b c = 0 ∴ {[a (b - c)]}^{3} + {[b (c - a)]}^{3} + {[c (a - b)]}^{3} = 3 [a (b - c)] [b (c - a)] [c (a - b)] = 3 a b c (b - c) (c - a) (a - b)$

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Please answer 19 i) and ii) Please answer 19 i) and ii) — — I yqo + P) + (91 ' KJ"duns ut •qj 10 q »eo s.'ojse,l JO tuns pull s.»qn» m.' Jumnut xuo•ut •BOI ve

Please answer 19 i) and ii)

Please answer 19 i) and ii) — — I yqo + P) + (91 ' KJ"duns ut •qj 10 q »eo s.'ojse,l JO tuns pull s.»qn» m.' Jumnut xuo•ut •BOI ve