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Q : If a + b + c = 0 , then prove that
![\frac{ {(b + c)}^{2} }{3bc} + \frac{ {(a + c)}^{2} }{3ac} + \frac{ {(a + b)}^{2} }{3ab} = 1 \frac{ {(b + c)}^{2} }{3bc} + \frac{ {(a + c)}^{2} }{3ac} + \frac{ {(a + b)}^{2} }{3ab} = 1](https://tex.z-dn.net/?f=+%5Cfrac%7B+%7B%28b+%2B+c%29%7D%5E%7B2%7D+%7D%7B3bc%7D++%2B+++%5Cfrac%7B+%7B%28a+%2B+c%29%7D%5E%7B2%7D+%7D%7B3ac%7D++%2B+++%5Cfrac%7B+%7B%28a+%2B+b%29%7D%5E%7B2%7D+%7D%7B3ab%7D++%3D+1)
For Calculation see pic
For Calculation see pic
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