Prove :1/a^{2n+1}+1/b^{2n+1}+1/c^{2n+1}=1/{a^{2n+1}+b^{2n+1}+c^{2n+1}}

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SUMMARY

The discussion centers on proving the mathematical identity $\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}$ given the condition $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$ for natural numbers $n$. Participants emphasize the importance of understanding the relationship between the terms and the implications of the identity in mathematical proofs. The discussion highlights the necessity of rigorous proof techniques in algebraic identities.

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Albert1
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$given$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$
$prove :$
$if \,\, n\in N\,\, then $
$\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}$
 
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Albert said:
$given$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$
$prove :$
$if \,\, n\in N\,\, then $
$\dfrac{1}{a^{2n+1}}+\dfrac{1}{b^{2n+1}}+\dfrac{1}{c^{2n+1}}=\dfrac{1}{a^{2n+1}+b^{2n+1}+c^{2n+1}}$

Albert and others Happy new year

we have
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\dfrac{1}{a+b+c}$
$=>\frac{1}{a}+\frac{1}{b} = \frac{1}{a+b+c}- \frac{1}{c}$
$=>\frac{a+b}{ab} = - \frac{a+b}{c(a+b+c)}$
$=>(a+b) = 0\cdots(1)$ or $\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$=>c(a+b+c) + ab = 0$
or $(a+c)(b+c) = 0$
so from (1) and above we have $a= -b$ or $b= -c$ or $c= - a$
beacuse of symmetry taking $a = -b$ we have a + b+ c = c and both LHS and RHS of given expression $\frac{1}{c^{2n+1}}$ and same
hence proved
 
kaliprasad said:
Albert and others Happy new year

we have
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\dfrac{1}{a+b+c}$
$=>\frac{1}{a}+\frac{1}{b} = \frac{1}{a+b+c}- \frac{1}{c}$
$=>\frac{a+b}{ab} = - \frac{a+b}{c(a+b+c)}$
$=>(a+b) = 0\cdots(1)$ or $\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$\frac{1}{ab} + \frac{1}{c(+b+c)} = 0$
$=>c(a+b+c) + ab = 0$
or $(a+c)(b+c) = 0$
so from (1) and above we have $a= -b$ or $b= -c$ or $c= - a$
beacuse of symmetry taking $a = -b$ we have a + b+ c = c and both LHS and RHS of given expression $\frac{1}{c^{2n+1}}$ and same
hence proved
Happy new year to all MHB folks
 

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