Question 138

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

Solution

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

Multiplying both sides by 2, we get :

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

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

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

$$\because$$ Sum of three positive sum is 0, then each term is equal to '0'

=> $$(a-b)=(b-c)=(c-a)=0$$

=> $$a=b=c$$

$$\therefore$$ $$a:b:c=1:1:1$$

=> Ans - (B)

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