Upside Down

$\dfrac 1a+\dfrac 1b+\dfrac 1c=0$
$(a+b+c)(\dfrac 1a+\dfrac 1b+\dfrac 1c)=0$
$\dfrac ab+\dfrac ba+\dfrac ac+\dfrac ca+\dfrac bc+\dfrac cb=-3 \cdots (1)$
$\dfrac 1a+\dfrac 1b+\dfrac 1c=0$
$\dfrac 1a=-\dfrac 1b-\dfrac 1c \cdots (2)$
$\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}$
$=(-\dfrac 1b-\dfrac 1c)^2(bc)+(-\dfrac 1a-\dfrac 1c)^2(ac)+(-\dfrac 1a-\dfrac 1b)^2(ab)) \text{ from }(2)$
$=6+\dfrac ab+\dfrac ba+\dfrac ac+\dfrac ca+\dfrac bc+\dfrac cb$
$=6-3 \text{ from }(1)$
$=3$

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0 \iff \frac{1}{c}=-(\frac{1}{a}+\frac{1}{b})$ $\iff \frac{1}{c^{3}}=-(\frac{1}{a}+\frac{1}{b})^{3}$ $\iff \frac{1}{c^{3}}=-(\frac{1}{a^{3}}+\frac{3}{a^{2}b}+\frac{3}{ab^{2}}+\frac{1}{b^{3}})=-\frac{1}{a^{3}}-\frac{1}{b^{3}}-(\frac{3}{a^{2}b}+\frac{3}{ab^{2}}) \qquad (*)$ On the other hand, $\frac{3}{a^{2}b}+\frac{3}{ab^{2}}$ $=\frac{1}{a} \times \frac{3}{ab}+\frac{1}{b} \times \frac{3}{ab}$ $=\frac{3}{ab}(\frac{1}{a}+\frac{1}{b})$ $=\frac{3}{ab} \times -\frac{1}{c}$ $=-\frac{3}{abc}$ $(*) \quad \iff \quad \frac{1}{c^{3}}=-\frac{1}{a^{3}}-\frac{1}{b^{3}}+\frac{3}{abc}$ $\iff \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}=\frac{3}{abc}$ So $\frac{bc}{a^{2}} + \frac{ac}{b^{2}} + \frac{ab}{c^{2}}$ $=\frac{a \times bc}{a \times a^{2}}+\frac{b \times ac}{b \times b^{2}}+\frac{c \times ab}{c \times c^{2}}$ $=\frac{abc}{a^{3}}+\frac{abc}{b^{3}}+\frac{abc}{c^{3}}$ $=abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$ $=abc \times \frac{3}{abc}$ $=\Large \boxed{3}$

$\dfrac{bc}{a^2} + \dfrac{ac}{b^2} + \dfrac{ab}{c^2}$

We know that:

$\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 0\\ \dfrac{bc+ac+ab}{abc}=0\\ ab+ac+bc=0$

This implies that:

$bc=-ab-ac \quad ac=-ab-bc \quad ab=-ac-bc$

Substitute these in:

$\dfrac{bc}{a^2} + \dfrac{ac}{b^2} + \dfrac{ab}{c^2}\\ =\dfrac{-ab-ac}{a^2} + \dfrac{-ab-bc}{b^2} + \dfrac{-ac-bc}{c^2}\\ =-\left( \dfrac{b+c}{a} + \dfrac{a+c}{b} + \dfrac{a+b}{c} \right)\\ =-\left( \dfrac{bc(b+c) + ac(a+c) + ab(a+b)}{abc} \right)\\ =-\left( \dfrac{b^2c + bc^2 + a^2c + ac^2 + a^2b + ab^2}{abc} \right)\\ =-\left( \dfrac{a^2b + a^2c + ab^2 + b^2c + bc^2 + ac^2}{abc} \right)\\ =-\left( \dfrac{a(ab + ac) + b(ab + bc) + c(bc + ac)}{abc} \right)$

Again, from the equation $ab+ac+bc=0$ , we know that:

$ab+ac=-bc \quad ab+bc=-ac \quad ac+bc=-ab$

Substitute again:

$-\left( \dfrac{a(ab + ac) + b(ab + bc) + c(bc + ac)}{abc} \right)\\ =-\left( \dfrac{a(-bc) + b(-ac) + c(-ab)}{abc} \right)\\ =\dfrac{3abc}{abc}\\ =\boxed{3}$

Substitute $(a,b,c)=(1,\omega,\omega^2)$ where $\omega$ denotes complex cube root of unity and then using $\omega^3=1$ the expression simplifies to $\boxed{3}$ .

There are 2 ways to prove this: 1. We will use $a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$ or $a^{3} + b^{3} + c^{3} = (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca) + 3abc$ (1) (try to prove it yourself, or there is a proof at the end of this solution). $\frac{bc}{a^{2}} + \frac{ac}{b^{2}} + \frac{ab}{c^{2}} \\ = \frac{abc}{a^{3}} + \frac{abc}{b^{3}} + \frac{abc}{c^{3}} \\ =abc\left ( \frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}} \right ) \\ =abc\left [ \left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right ) \left ( \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} - \frac{1}{ab} - \frac{1}{bc} - \frac{1}{ca} \right ) + 3\cdot \frac{1}{a}\cdot \frac{1}{b}\cdot \frac{1}{c} \right ] \\ =abc\cdot \frac{3}{abc} \\ =3$ 2. From $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$ , we got that: $\frac{1}{a} + \frac{1}{b} = -\frac{1}{c} \\ \left ( \frac{1}{a} + \frac{1}{b} \right)^{3} = -\frac{1}{c^{3}}$ Now, we have another fact that $a^{3} + b^{3}=(a + b)^{3} - 3ab(a+b)$ , you can prove this by expanding $(a+b)^{3}$ . Thus: $\frac{bc}{a^{2}} + \frac{ac}{b^{2}} + \frac{ab}{c^{2}} \\ =abc\left ( \frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}} \right ) \\ =abc\left [ \left ( \frac{1}{a} + \frac{1}{b} \right )^{3} - 3\frac{1}{ab}\left ( \frac{1}{a} + \frac{1}{b} \right ) + \frac{1}{c^{3}}\right ] \\ =abc\left [ -\frac{1}{c^{3}} - \frac{3}{ab}\left ( -\frac{1}{c} \right ) + \frac{1}{c^{3}} \right ] \\ =abc\cdot \frac{3}{abc} \\ =3$ Proof of (1): You may prove that $a^{3} + b^{3}=(a + b)^{3} - 3ab(a+b)$ . Using this, we have: $a^{3}+b^{3}+c^{3}-3abc \\ =(a+b)^{3}-3ab(a+b)+c^{3}-3abc \\ =(a+b)^{3}+c^{3}-3ab(a+b)-3abc \\ =(a+b+c)[(a+b)^{2}-(a+b)c+c^{2}]-3ab(a+b+c) \\ =(a+b+c)(a^{2}+2ab+b^{2}-ac-bc+c^{3}-3ab) \\ =(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

Let a, b and c be the roots of $x^3 - Sx^2 - abc=0$ , the coefficient of $x$ is zero (the sum of reciprocals being 0). Dividing this by $x^3$ we get the equation, $1 - \frac{S}{x} - \frac {abc}{x^3}=0$ . As the expression to be evaluated can be re-written as $abc(\frac{1}{a^3}+\frac {1}{b^3} + \frac {1}{c^3})$ , after putting x=a, then x=b, then x=c in the equation succesively and adding them gives $3 - S(\frac {1}{a} + \frac{1}{b} + \frac{1}{c}) - abc(\frac {1}{a^3} + \frac {1}{b^3} + \frac {1}{c^3})=0$ , where multiplier of S is 0 (given), we find the value of the required expression as 3. ANSWER