Question

If a+b+c=0 then prove that a3+b3+c3= 3abc , only when a=b=c

Answer 1

Step-by-step explanation:

a+b+c=0

we know that,

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

since, a+b+c=0

so,

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

=> a³+b³+c³-3abc = 0

=> a³+b³+c³ = 3abc

when a+b+c is not equal to 0,

we know that,

a³+b³+c³-3abc=1/2(a+b+c){(a-b)²+(b-c)²+(c-a)²}

To be a³+b³+c³ = 3abc, one of its factor must be 0

since a+b+c is not equal to 0

so, {(a-b)²+(b-c)²+(c-a)²} =0

so, a-b=0, b-c=0, c-a=0

so, a=b, b=c and c=a

i.e. a=b=c

Answer 2

Answer:

we know that

a^3+b^3 + c^3  = (a+ b + c) (a2 + b2 + c2 – ab – bc – ca) + 3abc

[using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)]

= 0 + 3abc [∴ a + b + c = 0]

a3+b3 + c3 = 3abc.

please mark it as brainliest

Step-by-step explanation: