If a + b + c = 0, then what is the value of \(\frac{a^2+b^2+c^2}{(a-b)

1.	If a + b + c = 0, then what is the value of \(\frac{a^2+b^2+c^2}{(a-b)^+(b-c)^2+(c-a)^2}\)(a) 1 (b) 3 (c) \(\frac{1}{3}\)(d) 0
Answer» (c) \(\frac13\) a + b + c = 0 ⇒ (a + b + c)2 = 0 ⇒ a² + b² + c² + 2ab + 2bc + 2ca = 0 ⇒ a² + b² + c² = – (2ab + 2bc + 2ca) Now, \(\frac{a^2+b^2+c^2}{(a-b)^+(b-c)^2+(c-a)^2}\) = \(\frac{a^2+b^2+c^2}{a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca}\) = \(\frac{a^2+b^2+c^2}{2(a^2+b^2+c^2)-(2ab+2bc+2ca)}\) = \(\frac{a^2+b^2+c^2}{2(a^2+b^2+c^2)-(a^2+b^2+c^2)}\) = \(\frac{a^2+b^2+c^2}{3(a^2+b^2+c^2)}\) = \(\frac13\)

If a + b + c = 0, then what is the value of \(\frac{a^2+b^2+c^2}{(a-b)^+(b-c)^2+(c-a)^2}\)(a) 1 (b) 3 (c) \(\frac{1}{3}\)(d) 0

Answer»

(c) \(\frac13\)

a + b + c = 0 ⇒ (a + b + c)2 = 0

⇒ a² + b² + c² + 2ab + 2bc + 2ca = 0

⇒ a² + b² + c² = – (2ab + 2bc + 2ca)

Now, \(\frac{a^2+b^2+c^2}{(a-b)^+(b-c)^2+(c-a)^2}\)

= \(\frac{a^2+b^2+c^2}{a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca}\)

= \(\frac{a^2+b^2+c^2}{2(a^2+b^2+c^2)-(2ab+2bc+2ca)}\)

= \(\frac{a^2+b^2+c^2}{2(a^2+b^2+c^2)-(a^2+b^2+c^2)}\)

= \(\frac{a^2+b^2+c^2}{3(a^2+b^2+c^2)}\) = \(\frac13\)