The Zero Constraint

Algebra Level 3

1 a 2 + b 2 − c 2 + 1 a 2 − b 2 + c 2 + 1 b 2 − a 2 + c 2 \large \dfrac1{a^2 + b^2 - c^2} + \dfrac1{a^2 - b^2 + c^2} + \dfrac1{b^2-a^2+c^2}

If a , b a,b and c c are non-zero such that a + b + c = 0 a+b+c=0 , find the value of the expression above.

-1 0 1 2

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Viki Zeta by Viki Zeta , 19, India

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3 solutions

Viki Zeta
Sep 1, 2016

a + b + c = 0 ⟹ a + b = − c ( a + b ) 2 = ( − c ) 2 a 2 + b 2 + 2 a b = c 2 a 2 + b 2 − c 2 = − 2 a b ... 1 also; b + c = − a b 2 + c 2 + 2 b c = a 2 b 2 − a 2 + c 2 = − 2 b c ... 2 and; c + a = − b c 2 + a 2 + 2 a c = b 2 a 2 − b 2 + c 2 = − 2 a c ... 3 Now, using 1 , 2 , 3 1 a 2 + b 2 − c 2 + 1 a 2 − b 2 + c 2 + 1 b 2 − a 2 + c 2 = 1 − 2 a b + 1 − 2 a c + 1 − 2 b c = a + b + c − 2 a b c = 0 − 2 a b c = 0 a + b + c = 0 \\ \implies a + b = -c \\ (a+b)^2 = (-c)^2 \\ a^2 + b^2 + 2ab = c^2 \\ a^2 + b^2 - c^2 = -2ab \text{ ... } \fbox{1}\\ \text{also;} \\ b+c = -a\\ b^2 + c^2 + 2bc = a^2 \\ b^2 - a^2 + c^2 = -2bc \text{ ... } \fbox{2}\\ \text{and;} \\ c+a = -b \\ c^2 + a^2 + 2ac = b^2 \\ a^2 - b^2 + c^2 = -2ac \text{ ... } \fbox{3} \\ \text{Now, using } \fbox{1}, \fbox{2}, \fbox{3} \\ \dfrac{1}{a^2+b^2-c^2} + \dfrac{1}{a^2-b^2+c^2} + \dfrac{1}{b^2-a^2+c^2} \\ = \dfrac{1}{-2ab} + \dfrac{1}{-2ac} + \dfrac{1}{-2bc} \\ = \dfrac{a + b + c}{-2abc} \\ = \dfrac{0}{-2abc} = 0

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If we solve it in this way, the answer cannot be zero

1/(a^2 + (b-c)(b+c) ) + 1/( (a-b)(a+b) + c^2) + 1/ ( (b-a)(b+a) + c^2)

=1/(a^2 + (b-c)(b+c) ) + 1/( (a-b)(a+b) + c^2) - 1/( (a-b)(b+a) + c^2)

=1/(a^2 + b^2 - c^2) + 0 = 1/(-2ab) Since , a^2 + b^2 - c^2 = -2ab

In the question it is given that a, b, c are non-zero.

So 1/(-2ab) has some integer value.

I am confused, so tell what is wrong in solving the question in this way?

Nashita Rahman - 4 years, 9 months ago

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Only values of a,b,c must be non zero. a,b,c can be integers. And even if they are outside out doesn't affect the condition. How did you put '-' in 2nd step?

Viki Zeta - 4 years, 9 months ago

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In the 2nd step, I took '-' common from (b-a) , so the minus sign came out it became (a-b)

Nashita Rahman - 4 years, 9 months ago

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@Nashita Rahman – ( a − b ) ( b − a ) = a b − a 2 − b 2 + a b = − b 2 − a 2 . (a-b)(b-a) = ab - a^2 - b^2 +ab = -b^2 - a^2. So, it would give 2ab than -2ab

Viki Zeta - 4 years, 9 months ago

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@Viki Zeta – You are not getting what I am trying to say

1/( c^2 + (a-b)(a+b) ) = - 1/( c^2 + (b-a)(b+a) ) using this I got the 2nd step!!

Nashita Rahman - 4 years, 9 months ago

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@Nashita Rahman – Simplify it. You won't get -2ab but − 1 2 a b \dfrac{-1}{2ab}

Viki Zeta - 4 years, 9 months ago

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@Viki Zeta – Yes we will get 1/(-2ab) , I never said that we will get -2ab. So if we solve the problem in this way, the answer cannot be zero as 'a' & 'b' are non-zero!

Nashita Rahman - 4 years, 9 months ago

@Viki Zeta – No it wouldn't. You typoed the last bit, should be − a 2 − b 2 + 2 a b -a^2-b^2+2ab .

Sharky Kesa - 4 years, 9 months ago

X = 1 a 2 + b 2 − c 2 + 1 a 2 − b 2 + c 2 + 1 b 2 − a 2 + c 2 = 1 a 2 + ( b − c ) ( b + c ) + 1 c 2 + ( a − b ) ( a + b ) + 1 b 2 + ( c − a ) ( c + a ) = 1 a 2 + ( b − c ) ( a + b + c 0 − a ) + 1 c 2 + ( a − b ) ( a + b + c 0 − c ) + 1 b 2 + ( c − a ) ( a + b + c 0 − b ) = 1 a 2 − a ( b − c ) + 1 c 2 − c ( a − b ) + 1 b 2 − b ( c − a ) = 1 a ( a − b + c ) + 1 c ( − a + b + c ) + 1 b ( a + b − c ) = 1 a ( a + b + c 0 − 2 b ) + 1 c ( a + b + c 0 − 2 a ) + 1 b ( a + b + c 0 − 2 c ) = − 1 2 ( 1 a b + 1 c a + 1 b c ) = − 1 2 ( a + b + c 0 a b c ) = 0 \begin{aligned} X & = \frac 1{a^2+b^2-c^2} + \frac 1{a^2-b^2+c^2} + \frac 1{b^2-a^2+c^2} \\ & = \frac 1{a^2+(b-c)(b+c)} + \frac 1{c^2+(a-b)(a+b)} + \frac 1{b^2+(c-a)(c+a)} \\ & = \frac 1{a^2+(b-c)(\cancel{a+b+c}^0-a)} + \frac 1{c^2+(a-b)(\cancel{a+b+c}^0-c)} + \frac 1{b^2+(c-a)(\cancel{a+b+c}^0-b)} \\ & = \frac 1{a^2 - a(b-c)} + \frac 1{c^2 - c(a-b)} + \frac 1{b^2 - b(c-a)} \\ & = \frac 1{a(a-b+c)} +\frac 1{c(-a+b+c)} + \frac 1{b(a+b-c)} \\ & = \frac 1{a(\cancel{a+b+c}^0-2b)} + \frac 1{c(\cancel{a+b+c}^0-2a)} + \frac 1{b(\cancel{a+b+c}^0-2c)} \\ & = -\frac 12 \left( \frac 1{ab} + \frac 1{ca} + \frac 1{bc} \right) \\ & = -\frac 12 \left( \frac {\cancel{a+b+c}^0}{abc} \right) \\ & = \boxed{0} \end{aligned}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution! +1

Viki Zeta - 4 years, 9 months ago
Rocco Dalto
Sep 15, 2016

Let a + b + c = 0 {\bf a + b + c = 0 }

a 2 + b 2 − c 2 = ( a + b + c ) ( a + b − c ) − 2 a b = − 2 a b {\bf a^2 + b^2 - c^2 = (a + b + c)(a + b - c) -2ab = -2ab }

a 2 − b 2 + c 2 = ( a + c + b ) ( a + c − b ) − 2 a c = − 2 a c {\bf a^2 - b^2 + c^2 = (a + c + b)(a + c - b) -2ac = -2ac }

b 2 − a 2 + c 2 = ( b + c + a ) ( b + c − a ) − 2 b c = − 2 b c {\bf b^2 -a^2 +c^2 = (b + c + a)(b + c - a) -2bc = -2bc }

⟹ − 1 2 ∗ ( 1 a b + 1 a c + 1 b c ) = {\bf \implies \frac{-1}{2} * (\frac{1}{ab} + \frac{1}{ac} +\frac{1}{bc}) = }

− 1 2 ∗ ( a b c 2 + a c b 2 + b c a 2 a 2 b 2 c 2 ) = {\bf \frac{-1}{2} * (\frac{abc^2 + acb^2 + bca^2}{a^2b^2c^2}) = }

( − a b c ) ∗ ( a + b + c ) 2 a 2 b 2 c 2 = {\bf \frac{(-abc) * (a + b + c)}{2a^2b^2c^2} = }

− ( a + b + c ) 2 a b c = 0 {\bf \frac{-(a + b + c)}{2abc} = 0 }

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