# Let n in N,n=2k+1,and 1/(a+b+c)=1/a+1/b+1/c. Show that 1/(a^n+b^n+c^n) = 1/(a^n) + (1)/(b^n) + (1)/(c^n) I have tried, but I don't get anything. Can you please give me a hint?

Let
Show that $\frac{1}{{a}^{n}+{b}^{n}+{c}^{n}}=\frac{1}{{a}^{n}}+\frac{1}{{b}^{n}}+\frac{1}{{c}^{n}}$
I have tried, but I don't get anything. Can you please give me a hint?
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tucetiw0
From the original equation, we get:
$abc=\left(a+b+c\right)\left(ab+bc+ca\right)$
which is equivalent to
${a}^{2}\left(b+c\right)+bc\left(b+c\right)+ab\left(b+c\right)+ca\left(b+c\right)=0$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(b+c\right)\left(a+c\right)\left(a+b\right)=0$
Then, obviously any one of the following must hold:
$a=-b\phantom{\rule{0ex}{0ex}}b=-c\phantom{\rule{0ex}{0ex}}c=-a$
In any case we can prove the equation
$\frac{1}{{a}^{n}+{b}^{n}+{c}^{n}}=\frac{1}{{a}^{n}}+\frac{1}{{b}^{n}}+\frac{1}{{c}^{n}}$
with odd $n$. Since if we take $a=-b$ we get
$\frac{1}{\left(-b{\right)}^{n}+{b}^{n}+{c}^{n}}=\frac{1}{\left(-b{\right)}^{n}}+\frac{1}{{b}^{n}}+\frac{1}{{c}^{n}}$
which is equivalent to
$\frac{1}{{c}^{n}}=\frac{1}{{c}^{n}}$
and this is true.....
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Liberty Page
Hint
At the first step, show that $\left(a+b\right)\left(a+c\right)\left(b+c\right)=0$
Next, show that two of the three numbers are opposite.
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