An algebra problem by Sai Venkatesh

Algebra Level 2

Given that a , b a,b and c c are real numbers such that a + b + c = 0 a+b+c=0 , compute 1 b 2 + c 2 − a 2 + 1 c 2 + a 2 − b 2 + 1 a 2 + b 2 − c 2 . \frac { 1 }{ { b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 } } +\frac { 1 }{ { c }^{ 2 }+{ a }^{ 2 }-{ b }^{ 2 } } +\frac { 1 }{ { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }.


The answer is 0.

Sai Venkatesh by Sai Venkatesh , 30, India

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5 solutions

It is given that a + b + c = 0 ⇒ a = − ( b + c ) , b = − ( c + a ) , c = − ( a + b ) a+b+c=0\quad \Rightarrow a = - (b+c), b=-(c+a), c = -(a+b) .

Now consider b 2 + c 2 − a 2 = b 2 + c 2 − [ − ( b + c ) ] 2 = b 2 + c 2 − ( b 2 + 2 b c + c 2 ) = − b c b^2+c^2-a^2 = b^2+c^2 - [-(b+c)]^2 = b^2+c^2 - (b^2+2bc+c^2) = -bc .

Therefore,

1 b 2 + c 2 − a 2 + 1 c 2 + a 2 − b 2 + 1 a 2 + b 2 − c 2 = − 1 2 b c − 1 2 c a − 1 2 a b \dfrac {1}{b^2+c^2-a^2 } + \dfrac {1}{c^2+a^2-b^2 } + \dfrac {1}{a^2+b^2-c^2 } = - \dfrac {1}{2bc} - \dfrac {1}{2ca} - \dfrac {1}{2ab}

= − 1 2 ( 1 b c + 1 c a + 1 a b ) = − 1 2 ( a + b + c a b c ) = − 1 2 ( 0 a b c ) = 0 = - \dfrac {1}{2} \left( \dfrac {1}{bc} + \dfrac {1}{ca} + \dfrac {1}{ab} \right) = - \dfrac {1}{2} \left( \dfrac {a+b+c}{abc} \right) = - \dfrac {1}{2} \left( \dfrac {0}{abc} \right) = \boxed{0}

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Mitali Mohile
Aug 17, 2014

a+b+c=0 therefore a, b and c=0 so 1/0+1/0+1/0=0

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G i v e n a + b + c = 0. ⇒ a = − ( b + c ) . ⇒ a 2 = b 2 + c 2 + 2 b c . . . ( 1 ) N o w , b 2 + c 2 − a 2 = b 2 + c 2 − ( b 2 + c 2 + 2 b c ) . . . . [ S u b s t i t u t i n g t h e v a l u e o f a f r o m ( 1 ) ⇒ b 2 + c 2 − a 2 = − 2 b c . S i m i l a r y f o r t h e o t h e r t w o e x p r e s s i o n s . [ ( c 2 + a 2 − b 2 ) & ( a 2 + b 2 − c 2 ) ] . N o w , 1 b 2 + c 2 − a 2 + 1 c 2 + a 2 − b 2 + 1 a 2 + b 2 − c 2 = − ( 1 2 b c 1 2 a c 1 2 a b ) . ⇒ − ( a 2 a b c + b 2 a b c + c 2 a b c ) = 0. Given\quad a+b+c=0.\\ \Rightarrow a=-(b+c).\\ { \Rightarrow a }^{ 2 }={ b }^{ 2 }+{ c }^{ 2 }+2bc\quad ...(1)\\ Now,\\ { b }^{ 2 }+{ c }^{ 2 }{ -a }^{ 2 }={ b }^{ 2 }+{ c }^{ 2 }-({ b }^{ 2 }{ +c }^{ 2 }+2bc).\quad \quad ...[Substituting\quad the\quad value\quad of\quad a\quad from\quad (1)\\ \Rightarrow { b }^{ 2 }+{ c }^{ 2 }{ -a }^{ 2 }=\quad -2bc.\\ Similary\quad for\quad the\quad other\quad two\quad expressions.\quad [({ c }^{ 2 }{ +a }^{ 2 }{ -b }^{ 2 })\quad \& \quad ({ a }^{ 2 }{ +b }^{ 2 }{ -c }^{ 2 })].\\ \\ Now,\\ \frac { 1 }{ { b }^{ 2 }+{ c }^{ 2 }{ -a }^{ 2 } } +\frac { 1 }{ { c }^{ 2 }{ +a }^{ 2 }{ -b }^{ 2 } } +\frac { 1 }{ { a }^{ 2 }{ +b }^{ 2 }{ -c }^{ 2 } } =\quad -(\frac { 1 }{ 2bc } \frac { 1 }{ 2ac } \frac { 1 }{ 2ab } ).\\ \Rightarrow -(\frac { a }{ 2abc } +\frac { b }{ 2abc } +\frac { c }{ 2abc } )=0.

Soummo Mukherjee - 6 years, 10 months ago

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Excellent. Why don't you post this as a solution here?

Krishna Ar - 6 years, 9 months ago

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exactly your solution is more accurate.

Mardokay Mosazghi - 6 years, 9 months ago

Exactly the same way Bro.

Kushagra Sahni - 6 years, 9 months ago

1/0=infinite....not zero

Saikarthik Bathula - 6 years, 9 months ago

a+b+c=0 therefore a,b and c=0 so 1/0+1/0+1/0=0

mitali mohile - 6 years, 10 months ago

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No its wrong because 1/0 is not defined. And also a, b, c are not necessarily =0. Only their sum is 0.

Kushagra Sahni - 6 years, 9 months ago

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Exactly a+b+c=0 always doesnt mean a,b & c are also 0. They may be negative integers

Satyajit Ghosh - 6 years, 9 months ago

mitali ur answer is write but steps are wrong any number by 0 is undefined

Sai Venkatesh - 6 years, 10 months ago
Mhar Ariz Marino
Dec 8, 2014

add the fractions. it will leave you with an a+b+c numerator which means 0. therefore the fraction is 0.

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Deepthi Prakash
Aug 22, 2014

a+b+c=0 = 0+0+0=0 0square is 0 (1/0)+(1/0)+(1/0)=0

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is this right sai venkatesh?

deepthi prakash - 6 years, 9 months ago

a+b+c=0...therefore...(a+b)^2 =c^2....=> a^2+b^2=c^2 -2ab.........therefore manipulating it you get it as zero...

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