Question

please answer for this

Answer 1

$using \: the \: identity \\ {(a + b + c)}^{3} = {a}^{3} + {b}^{3} + {c}^{3} + 3(a + b + c)(ab + bc + ca) - 3abc - - - (1) \\ let \: a = \frac{l}{m} . \: \: b = \frac{m}{n} . \: \: c = \frac{n}{l} \\ then \: \: given \: \: condition \: \: becomes \\ a + b + c = 0 - - - - (2) \\ also \\ abc = \frac{l}{m} \times \frac{m}{n} \times \frac{n}{l} = 1 - - - (3) \\ substituting \: (2) \: \: and \: \: (3) \: \: in \: identity (1) \\ {0}^{3} = {a}^{3} + {b}^{3} + {c}^{3} + 3 \times 0 \times (ab + bc + ca) - 3 \times 1 \\ thus \: \: {a}^{3} + {b}^{3} + {c}^{3} = 3 \\ therefore \\ \frac{ {l}^{3} }{ {m}^{3} } + \frac{ {m}^{3} }{ {n}^{3} } + \frac{ {n}^{3} }{ {l}^{3} } = 3$