Playing with Algebra
Let a , b and c be real numbers such that a = b and
a 2 ( b + c ) = b 2 ( a + c ) = 2
What is the value of c 2 ( a + b ) ?
The answer is 2.
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6 solutions
i think i have a solution
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Post it! :D
Yes post it! : )
Nice thinking. Congratulations.
i thinks its correct
Why and how did you go from first line to second line?
Amazing! Just curious though. What makes you think of multiplying (ab+ac+bc) with (a-c)? Where did you get (a-c)?
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I think i got the answer to my own question. I think it doesn't matter if we multiply it with (a-c) or (b-c). As long as we are trying to prove c2(a+b) will be equal to either of the two given expressions, it will just depend on which we choose. in the given solution, he used a-c so he came up with c2(a+b) being equal to the expression with "a" being squared. and if he opted to multiply it with b-c, it will then be equal to the one with "b" being squared. Sorry to sound unclear, but i already got your idea. thanks for this. it's a new magic trick for me... :-)
factoring of special products. . :)
how is your ac(a - c) + b(a-c)(a+c) = a^2(b+c) _ c^2(a+b)?
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a c ( a − c ) + b ( a 2 − b 2 ) = ( a 2 c + a 2 b ) − ( c 2 a + c 2 b ) , which is a 2 ( b + c ) − c 2 ( a + b ) = 0 . Sorry for not explain it in the solution, but I didn´t have enough time.
why did you multiply with (a-c) ?
After Calvin Lin explanation below I am adding this. \color\red{ a^2(b+c)- b^2(a+c) = 0~ and~ a~\neq~ b.\\gives~ ab+bc+ca=0, ~\therefore~a,b,c ~~are~~ cyclic.}
In a set of three numbers, two number follow the rule:- ( a n u m b e r ) 2 × ( s u m o f t h e r e m a i n i n g t w o ) = 2 . So applying this same rule to the third gives us c 2 ( a + b ) = 2 . 2
How do you know that the rule will definitely apply to the third case? (where c is the number being squared)
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Since this seems to be a case of cyclic variables.
After Calvin Lin explanation below I am adding this.
\color\red{ a^2(b+c)- b^2(a+c) = 0~ and~ a~\neq~ b.\\gives~ ab+bc+ca=0, ~\therefore~a,b,c ~~are~~ cyclic.}
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What's the proof?
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@Happy Melodies – I have given it now. See above.
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@Niranjan Khanderia – I disagree with your explanation which simply relies on cyclicity. For example, consider a similar question where a = b and a 2 b = b 2 c = 0 , that doesn't tell us anything about c 2 a even though it is cyclic. We could have b = 0 , while a and c are any non-zero value.
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@Calvin Lin – Thank you. I correct my thinking.
@Calvin Lin – I think that Cyclicity is valid when all the fuction have all variable .So here a^2b doesn't contain the variable c.
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@Ashish Rathor – In the specific example given, the fact that a 2 ( b + c ) = b 2 ( c + a ) , along with a = b , implies that the cyclic polynomial a b + b c + c a is equal to 0. This is why b 2 ( c + a ) = c 2 ( a + b ) .
However, this statement is not true in general. It is easy to find polynomials f ( a , b , c ) where f ( a , b , c ) − f ( b , c , a ) is not a cyclic polynomial.
For example, if
a 2 + b + c = b 2 + c + a = k .
We could have a = 0 , b = 1 , c = k − 1 , which gives us c 2 + a + b = k 2 − 2 k + 3 . In most cases, this would not be equal to k .
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@Calvin Lin
–
Thank you.
In
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is it because the starting points in both are different it may not be cyclic?
If we had
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would it be cyclic ?
I am greatly interested in cyclic functions involving more than two variables. Where can I go on the web?
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@Niranjan Khanderia – http://www.qc.edu.hk/math/Resource/AL/Cyclic%20and%20symmetric%20polynomials.pdf. This is a nice article on cyclic and symmetric functions I just found.
Write a comment or ask a question... This is not necessary.
Let a = 1, b = 1 + sqrt(3) and c = 1 - sqrt(3) as a found case.
c^2 (a + b) = 2 {By substitution}
At least you can know this ought to be true.
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Good thinking. Thanks.
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This is a short cut of particular case for verifying an identity.
Sqrt of (3) means?
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square root of 3. (square root of 4 is 2)
^_^ I use It !!!!!!
We notice that the variables really don't matter. Due to symmetry the answer is simply two.
why not -2 is the answer; i am getting c(a+b)=-ab ------(1) c^2(a+b)=-abc ------- * by c -------------(2) and c^2=4/((ab)^2) which give c=2/ab , -2/ab put in (2) we get answer -2,2
I have a lengthy solution
a2(b+c) = b2(a+c) = 2 => a2b + ca2 = ab2 + b2c => a2b - ab2 = b2c - ca2 => ab(a-b) = -c(a+b)(a-b) => -c = ab/(a+b) [as a!=b, so a-b can be divided from each hand] ------------- ---- (i) => c2 = a2b2 /(a+b)2 -------------------------------- (ii) => c2(a+b) = a2b2 /(a+b) ----------------------------- (iii) => c2(a+b) = a2(b+c) b2 (a+c) / ((a+b)(b+c)(c+a)) => c2(a+b) = 4 / ((a+b)(b+c)(c+a)) => c2(a+b) = 4 / ((b2c+ab2)+(a2b+ca2)+c2(a+b)+2abc) => x = 4/(2+2+x+2abc) [for simplicity, we put c2(a+b) = x; now we will find out value of x only] => x = 4/(4+x+2.ab.(-ab/(a+b))) [put the value of c from equation iii] => x = 4/(4+x-2a2b2/(a+b)) => x = 4/(4+x-2x) [derive the value of x from equation ii] => x = 4/(4-x) => -x2+4x = 4 => x2-4x+4=0 => (x-2)2 = 0 => x = 2 => c2(a+b) = 2
We have
0 = a 2 ( b + c ) − b 2 ( a + c ) = a b ( a − b ) + c ( a 2 − b 2 ) = ( a − b ) ( a b + a c + b c )
As a = b , we have a b + a c + b c = 0 . Multiplying by ( a − c ) we get
0 = ( a − c ) ( a b + a c + b c ) = a c ( a − c ) + b ( a 2 − c 2 ) = a 2 ( b + c ) − c 2 ( a + b )
in which c 2 ( a + b ) = a 2 ( b + c ) = 2 .