# If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ then $\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$

Suppose that $\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ . Then , prove that $\displaystyle\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}=\frac{1}{a^5+b^5+c^5}.$

Attempt :

From the given relation , $\displaystyle \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}=-2$.........(1)

Now I want to calculate , $\displaystyle \frac{a^5}{b^5}+\frac{b^5}{a^5}+\frac{b^5}{c^5}+\frac{c^5}{b^5}+\frac{a^5}{c^5}+\frac{c^5}{a^5}$. I tried by expanding pair of terms and putting the value of (1), but it can't help...

Hint:

$$(ab + bc + ca)(a + b + c) = (a + b)(b + c)(c + a) + abc.$$

Given $$\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}\Rightarrow \frac{(ab+bc+ca)}{abc} = \frac{1}{a+b+c}$$

So $$\displaystyle (ab+bc+ca)(a+b+c) = abc\Rightarrow a^2b+a^2c+ab^2+b^2c^2+c^2a+3abc=abc$$

so we get $(a+b)(b+c)(c+a) = 0\Rightarrow (a+b) = 0$ or $(b+c) =0$ or $(c+a) =0$

So $a=-b$ or $b=-c$ or $c=-a$

So $a=c$ or $b=-c$

So $$\displaystyle \frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5} = \frac{1}{c^5}$$ and $$\displaystyle \frac{1}{a^5+b^5+c^5} = \frac{1}{c^5}$$

So $$\displaystyle \frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5} = \frac{1}{a^5+b^5+c^5}$$

• Loved it +1 ! Nice work :) – user220382 Sep 8 '15 at 3:22