Question

if a,b,c are all non zero and a+b+c=0 prove that a^2/bc+b^2/ca+c^2/ab=3​

Answer 1

Answer:

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Answer 2

$\bf{\large{\underline{\underline{GIVEN:-}}}}$

a,b,c are all non zero

a+b+c=0

$\bf{\large{\underline{\underline{REQUIRED \: TO \: PROVE:-}}}}$

$\tt \dfrac{ {a}^{2} }{bc} + \dfrac{ {b}^{2} }{ca} + \dfrac{ {c}^{2} }{ab} = 3$

$\bf{\large{\underline{\underline{PROOF:-}}}}$

$\tt \dfrac{ {a}^{2} }{bc} + \dfrac{ {b}^{2} }{ca} + \dfrac{ {c}^{2} }{ab} = 3$

Consider LHS

$\tt = \dfrac{ {a}^{2} }{bc} + \dfrac{ {b}^{2} }{ca} + \dfrac{ {c}^{2} }{ab}$

Taking Least Common Multiple for the above terms

$\tt = \dfrac{ {a}^{2}(a)}{bc(a)} + \dfrac{ {b}^{2}(b)}{ca(b)} + \dfrac{ {c}^{2}(c)}{ab(c)}$

$\tt = \dfrac{ {a}^{3}}{abc} + \dfrac{ {b}^{3}}{abc} + \dfrac{ {c}^{3}}{abc}$

$\tt = \dfrac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc}$

$\boxed{ \large{ \sf when \: x + y + z = 0 \implies {x}^{3} + {y}^{3} + {z}^{3} = 3xyz}}$

And it is given a + b + c = 0

So a³ + b³ + c³ = 3abc

substitute a³ + b³ + c³ in the above fraction as 3abc

$\tt = \dfrac{3abc}{abc}$

$\tt = \dfrac{3 \cancel{abc}}{ \cancel{abc}}$

$\tt = \dfrac{3}{1}$

$\tt = 3$

$\tt = RHS \: i.e \: 3$

Hence Proved