Conditional Identities

Algebra Level 3

Also, a b c ≠ 0 abc \neq 0 . Solve it


The answer is 2.

Satyajit Ghosh by Satyajit Ghosh , 21, India

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3 solutions

We already have a common denominator, so we'll focus on the sum of the numerators. Since a + b + c = 0 a + b + c = 0 , the sum of the numerators can be rewritten as

( a + b ) 2 − a b + ( a + c ) 2 − a c = ( − c ) 2 + ( − b ) 2 − a ∗ ( b + c ) = b 2 + c 2 + ( b + c ) 2 = 2 ∗ ( b 2 + c 2 + b c ) (a + b)^2 - ab + (a + c)^{2} - ac = (-c)^{2} + (-b)^{2} - a*(b + c) = b^{2} + c^{2} + (b + c)^{2} = 2*(b^{2} + c^{2} + bc) .

Dividing this by the common denominator gives us an answer of 2 \boxed{2} .

9 Helpful 0 Interesting 0 Brilliant 0 Confused

But what if a,b,c are 0.

Anuj Shikarkhane - 6 years, 9 months ago

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Good point. If a , b , c a,b,c are all equal to 0 0 then the expression is indeterminate, so I took it as implicit that this wasn't the case. The asker should probably make this explicit in the question.

Note that if not all three are 0 0 then at least two must be non-zero, in which case at least one of b b or c c is non-zero. As long as one of them is non-zero then we are guaranteed that b 2 + c 2 + b c > 0 b^{2} + c^{2} + bc \gt 0 , in which case the expression will be determinate.

Brian Charlesworth - 6 years, 9 months ago

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Oh! I see. Thanks

Anuj Shikarkhane - 6 years, 9 months ago

Isn't it given clearly that abc is not eaual to zero, i.e. neither of them is zero.

Shivansh Nagi - 6 years, 5 months ago
Lee Isaac
Sep 20, 2015

Wow, the trick here is to see that since

a+b+c=0,

a+b=-c and b+c=-a and a+c=-b

This is an important step and afterwards the solution is obvious.

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I plugged values in for a,b,&c. Let a = .5 Let b = .3 Let c = -.8

Because a+b+c =0

When I plug those values in for the first fraction, the result is .49/.49 which is equal to 1.

When I plug those values in for the second fraction, the result is .49/.49 which is equal to 1.

1+1=2

There's your answer!

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let 1=1 b=2 c=-3

math man - 6 years, 9 months ago

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But the challenge is to solve it using variable

Satyajit Ghosh - 6 years, 9 months ago

If you plug in c = − a − b c = -a-b , it's not hard to see that both fractions are equal to 1 1 no matter which values you choose.

Patrick Corn - 6 years, 9 months ago

Ha Ha Ha ...... I did the same ^_^

Akshat Sharda - 5 years, 9 months ago

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