Question

If a+b+c=0 , then evaluate a3+b3+c3

Answer 1

Hey mate!

Here is yr answer......


Given,

a+b+c = 0


as we know,

a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c² -ab-bc-ac)

a³ + b³ + c³ -3abc = (0)(a²+b²+c²-ab-bc-ac)

a³+b³+c³ -3abc = 0

a³+b³+c³ = 3abc


Hope it helps....


#BeBrainly.

Answer 2

Your answer :-

$given = > a + b + c = 0 \\ find = > {a}^{3} + {b}^{3} + {c}^{3} \\ \\ by \: using \: identity.... \\ \\ = > {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca ) \\ \\put \: a + b + c = 0 \\ \\ = > {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (0)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca ) \\ \\ = > {a}^{3} + {b}^{ 3} + {c}^{3} - 3abc = 0 \\ \\ = > {a}^{3} + {b }^{3} + {c}^{3} = 3abc$

Thanks!!