Question

If $a,b,c$ are real and $a^3+b^3+c^3=3abc$ and $a+b+c\ne 0$, then the relation between $a,b,c$ will be:

• A. $a+b=c$
• B. $a+c=b$
• C. $a=b=c$
• D. $b+c=a$

Answer 1

The correct option is C $a=b=c$

if $a,b,c$ are real, $a^3+b^3+c^3=3abc$

given $a+b+c\ne 0$

Relation b/w $a,b,c=?$

$a^3+b^3+c^3=3abc$

$a^3+b^3+c^3-3abc=0$

fartrise the given expression

$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$

Given, $a+b+c\ne 0$

So, $a^2+b^2+c^2-ab-bc-ca=0$ (Must be, AS the product is $\Delta$)

Therefore,

$(a^2+b^2+c^2-ab-bc-ca)=0$

Multiply by 2 on both the sides

$(2a^2+2b^2+2c^2-2ab-2bc-2ca)=0$

Seprate individual square

$a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca=0$

$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$

$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$

We know $(a-b{)}^2,(b-c{)}^2,(c-a{)}^2>0$

So, for Add equal to 0, individual becomes 0

As $a-b=0$

$a=b...\left(1\right)$

As $b-c=0$

$b=c...\left(2\right)$

from equation (1) & (2)

$a=b=c$