Question

If $a^2+b^2+c^2-ab-bc-ca=0,$ then the value of a:b:c is

• A. 1:1:1
• B. 1:2:3
• C. 1:3:4
• D. 1:4:1

Answer 1

The correct option is A 1:1:1

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

Multiply both side by 2

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

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

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

Since the sum of square is zero then each term should be zero

$\Rightarrow (a-b{)}^2=0,(b-c{)}^2=0,(c-a{)}^2=0$

$\Rightarrow a-b=0,b-c=0,c-a=0$

$\Rightarrow a=b,b=c,c=a$

Then a=b=c

then ratio a:b:c=1:1:1