We know
then
Now,
Solution:−
\tt \: If \: a \: + \: b \: + \: c \: = \: 0Ifa+b+c=0
\tt \: {a}^{3} + {b}^{3} + {c}^{3} = 3abca
3
+b
+c
=3abc
\tt \: \dfrac{ {a}^{2} }{bc} + \dfrac{ {b}^{2} }{ca} + \dfrac{ {c}^{2} }{ab}
bc
a
2
+
ca
b
ab
c
\tt \: = \dfrac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc}=
abc
\tt \: = \dfrac{3abc}{abc}=
3abc
=3
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Answers & Comments
We know
then
Now,
Solution:−
We know
\tt \: If \: a \: + \: b \: + \: c \: = \: 0Ifa+b+c=0
then
\tt \: {a}^{3} + {b}^{3} + {c}^{3} = 3abca
3
+b
3
+c
3
=3abc
Now,
\tt \: \dfrac{ {a}^{2} }{bc} + \dfrac{ {b}^{2} }{ca} + \dfrac{ {c}^{2} }{ab}
bc
a
2
+
ca
b
2
+
ab
c
2
\tt \: = \dfrac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc}=
abc
a
3
+b
3
+c
3
\tt \: = \dfrac{3abc}{abc}=
abc
3abc
=3