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Question Number 187858 by normans last updated on 23/Feb/23

	Answered by som(math1967) last updated on 23/Feb/23
	$(1/a)+(1/b)+(1/c)=(1/(a+b+c)) ⇒(a+b+c)(ab+bc+ca)−abc=0 ⇒a^2 b+abc+a^2 c+ab^2 +b^2 c+abc+abc +bc^2 +c^2 a−abc=0 ⇒ab(a+c)+b^2 (a+c)+ca(a+c)+bc(a+c)=0 (a+c)(ab+b^2 +ca+bc)=0 (a+c){b(a+b)+c(a+b)}=0 ⇒(a+c)(a+b)(b+c)=0 if (a+c)=0∴a=−c (1/a^5 )+(1/b^5 )+(1/c^5 )=−(1/c^5 )+(1/b^5 )+(1/c^5 )=(1/b^5 ) again (1/((a+b+c)^5 ))=(1/b^5 ) if a+b=0⇒a=−b orb=−c gives (1/a^5 )+(1/b^5 )+(1/c^5 )=(1/((a+b+c)^5 ))$
	$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a + b + c}$ $\Rightarrow (a + b + c) (a b + b c + c a) - a b c = 0$ $\Rightarrow a^{2} b + a b c + a^{2} c + {a b}^{2} + b^{2} c + a b c + a b c$ $+ {b c}^{2} + c^{2} a - a b c = 0$ $\Rightarrow a b (a + c) + b^{2} (a + c) + c a (a + c) + b c (a + c) = 0$ $(a + c) (a b + b^{2} + c a + b c) = 0$ $(a + c) {b (a + b) + c (a + b)} = 0$ $\Rightarrow (a + c) (a + b) (b + c) = 0$ $i f (a + c) = 0 ∴ a = - c$ $\frac{1}{a^{5}} + \frac{1}{b^{5}} + \frac{1}{c^{5}} = - \frac{1}{c^{5}} + \frac{1}{b^{5}} + \frac{1}{c^{5}} = \frac{1}{b^{5}}$ $a g a i n \frac{1}{{(a + b + c)}^{5}} = \frac{1}{b^{5}}$ $i f a + b = 0 \Rightarrow a = - b o r b = - c$ $g i v e s \frac{1}{a^{5}} + \frac{1}{b^{5}} + \frac{1}{c^{5}} = \frac{1}{{(a + b + c)}^{5}}$
	Commented by normans last updated on 23/Feb/23

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