Question

If a+b+c=0 then prove that a³+b³+c³=3abc​

Answer 1

Answer:-

We know,

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

Given that: (a + b + c) = 0.

Putting the value:-

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

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

→ a³ + b³ + c³ - 3abc = 0

→ a³ + b³ + c³ = 3abc

Hence, Proved.

Answer 2

To prove:-

•If a+b+c=0,then a³+b³+c³=3abc

Solution:-

=>a+b+c=0

=>a+b= -c

=>(a+b)³=(-c)³

=>a³+b³+3ab(a+b)=-c³

=>a³+b³+3ab(-c)=-c³ (Since,a+b=-c)

=>a³+b³-3abc= -c³

=>a³+b³+c³=3abc

Thus,proved that-----

If (a+b+c)=0, then (a³+b³+c³)=3abc