Prove 1 a </mfrac> + 1 b </mfrac> + 1 c </mfrac> =

or5a2dosz80z

or5a2dosz80z

Answered question

2022-06-02

Prove 1 a + 1 b + 1 c = 1 a + b + c ( a + b ) ( b + c ) ( c + a ) = 0 without expansion

Answer & Explanation

lymbornenottemhr53

lymbornenottemhr53

Beginner2022-06-03Added 4 answers

p ( x ) = ( x a ) ( x b ) ( x c ) = x 3 u x 2 + v x w has roots a , b , c with sum a + b + c = u
( a + b ) ( b + c ) ( c + a ) = 0 ( u c ) ( u a ) ( u b ) = 0 p ( u ) = 0 thus:
u 3 u u 2 + v u w = 0 v w = 1 u 1 a + 1 b + 1 c = 1 a + b + c
Estrella Le

Estrella Le

Beginner2022-06-04Added 1 answers

We first start with a little rearrangement that is :
1 a + 1 b + 1 c = 1 a + b + c
implies :
( a b + b c + a c ) ( a + b + c ) a b c = 0
Now note try setting a = 0 we see then
b c ( b + c ) = 0
So In order to make this vanish set ,
b = c
But also observe that , this belongs to the solution set even if
a 0
that is :
( a ( b ) + b ( b ) + a ( b ) ) ( a + b b ) a b ( b ) = 0
Hence by symmetry the other roots must be
a = b , c = a
a = b , c = a
Therefore we have that :
( a + b ) ( b + c ) ( c + a ) = 0

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