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If $a+b+c=0$ and $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 3$. where $a,b,c$ are non zero real numbers

then $ab(a+b)+bc(b+c)+ca(c+a)$ is

Attempt: from $a+b+c=0$ we have to find value of $-3abc$

and from $\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 3$ we have $\displaystyle \frac{ab+bc+ca}{abc} = 3$

could some help how to get value of $abc,$ thanks

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We have $$(a+b+c)^2 =a^2+b^2+c^2+2 (ab+bc+ca) =0$$ $$\Rightarrow ab+bc+ca = -0.5 (a^2+b^2+c^2) $$

We already know $ab + bc + ca =3abc $, so hope you can take it from here.

On a side note, observe that $$ (a+b+c)^3 =a^3+b^3+c^3+ (a+b+c)(ab+bc+ca) \Rightarrow a^3+b^3+c^3=0$$

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    $\begingroup$ i have got $$\frac{3a^3}{3a-1}$$ what should be the result? $\endgroup$ – Dr. Sonnhard Graubner Feb 24 '17 at 10:33

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