Question

if a+b+c=0,then prove that a^2/bc+b^2/ca+c^2/ab

Answer 1

Hey There!!

Here, we are given one data:
a + b + c = 0


Here, to get the answer, we will use only one (not so) simple identity:
$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Let us solve your question:

$\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} \\ \\ \\ = \frac{a^3+b^3+c^3}{abc} \\ \\ \\ = \frac{a^3+b^3+c^3-3abc+3abc}{abc} \\ \\ \\ = \frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{abc} + \frac{3abc}{abc}} \\ \\ \\ = \frac{0\times (a^2+b^2+c^2-ab-bc-ca)}{abc} + 3 \\ \\ \\ = 0 + 3 \\ \\ \\ = 3$



Thus we have:
$\boxed{\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = 3}$

Hope it helps
Purva
Brainly Community

Answer 2

Answer:

Step-by-step explanation: