Question

hey friends,

if a+b+c=0, show that
${a}^{3} + {b}^{3} + {c}^{3} = 3abc$

Answer 1

hey mate,

$= a + b + c = 0 \\ = a + b = \: - 3 \\ = {(a + b)}^{3} = ({ - c})^{3} \\ = {a}^{3} + {b}^{3} + 3ab(a + b) = { - c}^{3} \\ = {a}^{3} + {b}^{3} + 3ab( - c) = { - c}^{3} \\ = {a}^{3} + {b}^{3} + {c}^{3 } - 3abc = 0 \\ = {a}^{3} + {b}^{3} + {c}^{3} = 3abc$


I hope this helps you!
@vaibhav246