Two Symmetric Equations

Simplifying the second expression: $\begin{array} {rcl} a\big(\tfrac{1}{b} + \tfrac{1}{c}\big) + b\big(\tfrac{1}{c} + \tfrac{1}{a}\big) + c\big(\tfrac{1}{a} + \tfrac{1}{b}\big) & = & -3 \\ a^2(b+c) + b^2(a + c) + c^2(a+b) & = & -3abc \\ a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2 + 3abc & = & 0 \\ (a+b+c)(ab+ac+bc) & = & 0 \\ \tfrac{1}{2}(a+b+c)\big[(a+b+c)^2 - (a^2+b^2+c^2)\big] & = & 0 \\ (a+b+c)((a+b+c)^2-1)& = & 0 \end{array}$ and hence $a+b+c$ can only take three values, namely $-1,0,1$ . Putting $a=\tfrac{1}{\sqrt{6}}$ , $b=\tfrac{1}{\sqrt{6}}$ , $c = -\tfrac{2}{\sqrt{6}}$ shows that $a+b+c=0$ is possible. Putting $a=-\tfrac13$ , $b=c=\tfrac23$ shows that $a+b+c=1$ is possible. Putting $a=\tfrac13$ , $b=c=-\tfrac23$ shows that $a+b+c=-1$ is also possible, Thus there are indeed three possible values for $a+b+c$ .

The equations are symmetric in $a,b,c$ , and so it is worthwhile investigating what the equations say in terms of the fundamental symmetric polynomials $a+b+c$ , $ab+ac+bc$ and $abc$ . Once we start on that line, the solution method is evident.

We can write the second equation as $a \cdot (\frac {1}{a} + \frac {1}{b} + \frac {1}{c} ) + b \cdot (\frac {1}{a} + \frac {1}{b} + \frac {1}{c} ) + c \cdot (\frac {1}{a} + \frac {1}{b} + \frac {1}{c} )$ =0 which will be reduced to $(a+b+c) \cdot (\frac {1}{a} + \frac {1}{b} + \frac {1}{c} ) = 0$ .

Therefore, either $(a+b+c) = 0$ or $(\frac {1}{a} + \frac {1}{b} + \frac {1}{c} ) = 0$ .

(1) One of the sets that satisfy the 2 equations when $a+b+c = 0$ is $\frac {1}{\sqrt{6}}, \frac {1}{\sqrt{6}}, -\frac {2}{\sqrt{6}}$ . So $a+b+c = 0$ is one solution.

(2) If $(\frac {1}{a} + \frac {1}{b} + \frac {1}{c} ) = 0$ , then $\frac {bc + ca + ab} {abc} = 0$ . So $bc + ca + ab$ = 0.

Now we look at $(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$ . Since $2ab + 2bc + 2ca$ = 0 and $a^2 + b^2 + c^2$ = 1, $(a+b+c)^2 = 1$ , $a+b+c = 1$ or $-1$ .

One set that satisfy the 2 equations when $a+b+c = -1$ is $-\frac {2}{3}, -\frac {2}{3}, \frac {1}{3}$ One set that satisfy the 2 equations when $a+b+c = 1$ is $\frac {2}{3}, \frac {2}{3}, -\frac {1}{3}$ So there are 2 more solutions.

In total, we have 3 solutions.

From the equation, $a(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+b(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+c(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) = 0$ Hence, $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=0$

Case 1: $a+b+c \not= 0$

Therefore, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ $(a+b+c)^2 - (a^2+b^2+c^2) = 2ab+2bc+2ac$ $= 2abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) = 0$ $a^2+b^2+c^2=1 \Rightarrow (a+b+c)^2=1 \Rightarrow a+b+c = 1 \text{ or }-1$

Case 2: $a+b+c = 0$

Therefore, there are 3 possible values of $a+b+c$ which are $-1, 0, 1$

Take the LCM(of the LHS) of the terms in the second equation:

$a(\frac{b+c}{bc})+b(\frac{a+c}{ac})+c(\frac{a+b}{ab})$

$=\frac{a^2(b+c)+b^2(a+c) + c^2(a+b)}{abc}$

Now, group the terms in the numerator such that we can bring the first equation into play:

$= \frac{a^2b+c^2b + ab^2+ac^2+ ca^2+cb^2}{abc}$

$=\frac{b(a^2+c^2)+a(b^2+c^2+c(a^2+b^2)}{abc}$

This is where we can use $a^2+b^2+c^2=1$ to simplify the expression obtained above.

$\frac{b(1-b^2)+a(1-a^2)+c(1-c^2)}{abc}$

$=\frac{(a+b+c)-(a^3+b^3+c^3)}{abc}$

The above expression equals $-3$ , and we have:

$(a+b+c)-(a^3+b^3+c^3)=-3abc$

$\Rightarrow (a+b+c)-(a^3+b^3+c^3-3abc)=0$

Use the well known factorization, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Take $(a+b+c)$ as the common factor:

$\Rightarrow (a+b+c)(1-(a^2+b^2+c^2-ab-bc-ca)=0$

$\Rightarrow (a+b+c)(ab+bc+ba)=0$

If $ab+bc+ca=0$

$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=1$

$\Rightarrow (a+b+c)^2=1$

Hence, we conclude that there are $\boxed{3}$ possible solutions for $a+b+c$ , namely : $0,1,-1$

We know that $\dfrac{a}{b}+ \dfrac{a}{c}+ \dfrac{b}{a}+ \dfrac{b}{c}+ \dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{c}{c}=0$ or $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=0.$

If $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0$ then $\dfrac{ab+bc+ca}{abc}=ab+bc+ca=0$ . Then $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=a^{2}+b^{2}+c^{2}=1.$ Therefore $a+b+c=\pm 1$ . On the other hand we may have $a+b+c=0$ as well.

So $a+b+c =0, -1, 1.$ It can be seen that: there are $a, b, c$ such that $a+b+c =0, -1$ or $1$ and $a^{2}+b^{2}+c^{2}=1$ and these solutions work.

The relation is equivalent to: $\dfrac{a}{b}+ \dfrac{a}{c}+ \dfrac{b}{a}+ \dfrac{b}{c}+ \dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{c}{c}=0$ or $(a+b+c)(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})=0.$ If $(a+b+c)=0$ we get one solution. If $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0$ then $\dfrac{ab+bc+ca}{abc}=ab+bc+ca=0$ . Then $(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca)=a^{2}+b^{2}+c^{2}=1\Rightarrow (a+b+c)=\pm 1$ .

So $a+b+c$ =0, -1, 1.) It can be seen that these solutions work and exists $a,b,c$ such that $a+b+c$ =0, -1 or 1 and $a^{2}+b^{2}+c^{2}=1$

Note that $a(\frac{1}{b}+\frac{1}{c})=a(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-1$ , so $a(\frac{1}{b}+\frac{1}{c})+b(\frac{1}{a}+\frac{1}{c})+c(\frac{1}{a}+\frac{1}{b})=a(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+$ $b(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+c(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3=(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-3$ $=-3$ , and adding three to both sides gives $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=0$ . By the zero-product property, either $a+b+c=0$ or $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ . Thus $a+b+c=0$ is one solution for $a+b+c$ . Otherwise, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc}{abc}+\frac{ac}{abc}+\frac{ab}{abc}=\frac{ab+ac+bc}{abc}=0$ . Since a, b, and c are non-zero, this is not undefined, and we can multiply by $abc$ to get $ab+ac+bc=0$ . Now, note that $(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$ Since $a^2+b^2+c^2=1$ , and $ab+ac+bc=0$ , $(a+b+c)^2=1+2(0)=1$ , so $a+b+c=1 or -1$ . Thus there are three possibilities for $a+b+c$ : -1, 0, and 1

Refer to a^2+b^2+c^2=1 as eqn 1 and a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=−3 as eqn 2.

we see that by adding the terms a/a=1 b/b=1 and c/c=1 to eqn 2, we can factorise it into the form (a+b+c)*(1/a+1/b+1/c)=0

This then gives us our first solution, a+b+c=0

we now consider the second case, where the reciprocals sum up to 0 and a+b+c \neq 0

call 1/a+1/b+1/c=0 eqn 3. We now multiple eqn 3 by abc, to get

ab+bc+ac=0, which we call eqn 4.

adding eqn 4 twice to eqn 1, we find that

a^2+b^2+c^2+2ab+2bc+2ac=1 which can be factorized into (a+b+c)^2=1

This then gives us our other 2 answers, a+b+c=1 and a+b+c=-1

In the second equation, get rid the denominator and we get: $a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2 = -3abc$ ; $a^2b + a^2c + ab^2 + b^2c + ac^2 + bc^2 +3abc = 0$ ; $(a^2b+ab^2 + abc) + (ac^2 + a^2c + abc) + (bc^2 + b^2c +abc)=0$ ; $ab(a+b+c) + ac(a+b+c) + bc(a+b+c) = 0$ ; $(ab+bc+ac)(a+b+c)=0$

Then we get: $a+b+c = 0$ or $ab+bc+ac=0$

Now consider case if $ab+bc+ac = 0$ , then from the first equation , it is equivalent to: $(a+b+c)^2 - 2(ab+bc+ac) = 1$ , because $ab+bc+ac = 0$ , so $(a+b+c)^2 = 1$

Then we get: $a+b+c = 1$ or $a+b+c = -1$

The conclusion is: the number of possible values for $a+b+c$ is $\underline{3}$ , which is $-1, 0 , 1$

consider the second expression,

it can be simplified as a^2(b+c)+b^2(a+c)+c^2(a+b)=-3abc;

which can be further simplified as;

(a+b+c)*(ab+bc+ac)=0;

ab+bc+ac=((a+b+c)^2-(a^2+b^2+c^2))/2;

considering a+b+c=x; x (x^2-1)/2=0; =>x (x^2-1)=0; x=0;x=1;x=-1; hence a+b+c can take three values

$\text{Multiplying by }abc,\\ a(\frac{1}{b}+\frac{1}{c})+b(\frac{1}{a}+\frac{1}{c})+c(\frac{ 1}{a}+\frac{1}{b})=-3\\ \Rightarrow a^2(b+c)+b^2(a+c)+c^2(a+b)=-3abc\\ \text{Expanding, we get}\\ a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+3abc=0\\ \Rightarrow (a+b+c)(ab+bc+ac)=0.\\ \text{We see that } a+b+c=0 \text{ is a solution.}\\ \text{For }(ab+bc+ac)=0,\\ \text{Notice that }(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac.\\ \text{Using our result and } a^2+b^2+c^2=1 \text{, We get } \\ (a+b+c)^2=1+2(ab+bc+ac)=1\\ \Rightarrow a+b+c=\pm1.\\ \text{Therefore, there are } 3 \text{ possible values {-1,0,1} of } a+b+c.$

-3=a(b+c/bc)+b(a+c/ac)+c(a+b/ab)=ab+ac/bc+ab+bc/ac+ac+bc/ab -3={(a^2 b)+(a^2 c)+(a b^2)+(b^2 c)+(a c^2)+(b c^2)}/abc -3={b(a^2+c^2)+c(a^2+b^2)+a(b^2+c^2)}/abc -3={b(1-b^2)+c(1-c^2)+a(1-a^2)}/abc. -3abc=(a^3+b^3+c^3+a+b+c) a+b+c=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) a^2+b^2+c^2=ab+bc+ca\ a=b=c. by solving according to this we can find 3 possible values for a,b,c.

a/b+b/a+c/a+a/c+b/c+c/b=-3 [a^2+b^2]/ab + [b^2+c^2]/bc +[c^2+a^2]/ac=-3 [1-c^2]/ab +[1-b^2]/ac +[1-a^2]/bc=-3 {a+b+c-[a^3+b^3+c^3]}/abc=-3 a+b+c=3abc+a^3+b^3+c^3 (a+b+c)=(a+b+c)(a^2+b^2+c^2+ab+bc+ca) (a+b+c)(ab+bc+ca)=0 [a^2+b^2+c^2=1] a+b+c=0 or ab+bc+ca=o i.e, (a+b+c)^2=a^2+b^2+c^2 (a+b+c)^2=1 a+b+c=1 or -1 i.e, a+b+c=0 or 1 or-1 3 solutions

a(1/b+1/c)+b(1/a+1/c)+c(1/b+1/a)=-3 a((b+c)/bc)+b((a+c)/ac)+c((a+b)/ab)=-3 [a^2(b+c)+b^2(a+c)+c^2(a+b)]/abc=-3 a^2b+a^2c+b^2a+b^2c+c^2a+c^2b=-3abc Adding a^3+b^3+c^3 to both sides, we get: a^3+b^3+c^3+a^2b+a^2c+b^2a+b^2c+c^2a+c^2b=a^3+b^3+c^3-3abc Factorising, (a^2+b^2+c^2)(a+b+c)=(a+b+c)(a^2+b^2+c^2-ab-ac-bc) Then if a+b+c=0, we are done. Otherwise, we get ab+bc+ca=0 a^2+b^2+c^2+2(ab+bc+ca)=1 (a+b+c)^2=1 so (a+b+c)=1 or -1 Thus, a+b+c can only be -1,0,1 when a,b,c are real numbers. Then we construct cases to show that it is possible to have a+b+c=-1,0 and 1.

First, let the LHS of the equation expands itself: $a (\frac{1}{b} + \frac{1}{c}) + b (\frac{1}{c} + \frac{1}{a}) + c (\frac{1}{a} + \frac{1}{b}) = (a+b+c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) - (\frac{a}{a} + \frac{b}{b} + \frac{c}{c}) = (a+b+c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) - 3$

So, with the value from the RHS, the problem now becomes:

$(a+b+c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = 0$

From above equation, the solution is either:

1. $a + b + c = 0$ or

2. $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}=0$

From 1 we get 1 solution.

From 2:

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

As $a, b, c$ are non zero real numbers, then $abc \neq 0$ , so $ab+ac+bc$ must be $0$

Because ( $a+b+c)^2 = a^2 + b^2+c^2 - 2(ab+ac+bc)$ , then ( $a+b+c)^2 = 1$ From this, we get $a+b+c = \pm 1$

So there are $\boxed{3}$ possible values of $a+b+c$ that hold that equation.

At first, the two given equations will seem completely unrelated but when we expand the second, we get $\frac ab + \frac ac + \frac bc + \frac ba + \frac ca + \frac bc = - 3.$ If we try to factor this, we know that this will have a factor with $a,b,c$ in the numerator and a factor with $a,b,c$ in the denominator. This seems to look like $(a+b+c)(\frac 1a + \frac1b + \frac1c) = 3 + \frac ab + \frac ac + \frac bc + \frac ba + \frac ca + \frac bc = 3 - 3 = 0.$ So we have two cases: $a+b+c = 0$ or $\frac 1a + \frac 1b + \frac1c= 0$ . The first case gives us one of the solution. If we were to use the second case, we see $\frac 1a + \frac 1b + \frac 1c = \frac{ab + bc + ac}{abc} = 0$ so $ab + bc + ac = 0.$ Now we use the fact that $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)$ to get $(a+b+c)^2 = 1 + 2(0) = 1.$ Thus we have two other solutions $a+b+c =1$ and $-1$ . This gives us a total of $3$ solutions.

$a \left( \frac{1}{b} + \frac{1}{c} \right) + b \left( \frac{1}{c} + \frac{1}{a} \right) + c \left( \frac{1}{a} + \frac{1}{b} \right) = -3$

Multiply both side with $abc$ then we will get:

$(a + b + c)(ab + bc + ca) = 0$

$\text{First Case:}$

$a + b + c = 0$

$\text{Second Case:}$

$(a + b + c)^2 - 2(ab + bc + ca) = 1$

$a + b + c = ± 1$

Hence, there are $\boxed{3}$ possible values for $a + b + c.$

The second equation is a(1/b + 1/c) + b(1/c + 1/a) + c(1/a + 1/b)= -3

Or, a(b+c)/bc + b(c+a)/ca + c(a+b)/ab= -3

Or, a^2(b+c) + b^2(c+a) + c^2(a+b) + 3abc = 0 [multiplying both sides by abc and using the fact that a,b,c are non zero numbers]

Or, (a+b)(b+c)(c+a) - 2abc + 3abc = 0 [since (a+b)(b+c)(c+a) = a^2(b+c) + b^2(c+a) + c^2(a+b) + 2abc]

Or, (a+b)(b+c)(c+a) + abc= 0

Now let p= a+b+c.

Then the equation becomes (p-a)(p-b)(p-c) + abc= 0

Or, p^3 - (a^2 + b^2 + c^2)p + (ab+bc+ca)p - abc + abc = 0 [when we expand (p-a)(p-b)(p-c) we see that (p-a)(p-b)(p-c) = p^3 - (a^2 + b^2 + c^2)p + (ab+bc+ca)p]

Or, p^3 - p + (ab+bc+ca)p = 0 [since a^2 + b^2 + c^2 =1]

Now note that (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca) Or, (ab+bc+ca) = (p^2 - 1)/2

Putting this value in the equation...

p^3 - p +(p^2 - 1)p/2= 0

Or, 3p^3 - p = 0

Or, p(3p^2 - 1)= 0

So, the possible values of p are...

i) p= 0 ii) p= sqrt(1/3) iii) p=-sqrt(1/3) [since sqrt(1/3) and -sqrt(1/3) are roots to the equation 3p^2 - 1 = 0]

This implies that there are 3 possible values of a+b+c.

The first thing to do is notice that:

$a\left(\frac{1}{b} + \frac{1}{c}\right) + b\left(\frac{1}{c} + \frac{1}{a}\right) + c\left(\frac{1}{a} + \frac{1}{b}\right) = \frac{a^2b+a^2c + ab^2+b^2c + ac^2+bc^2}{abc}$

This implies:

$a^2b + a^2 c + ab^2 + b^2c + ac^2 + bc^2 + 3abc = 0$

This expression can be factored into:

$(a+b+c)(ab + bc + ca) = 0$

Now if we look at:

$(a+b+c)^3 = (a+b+c)(a^2 + b^2 + c^2) + 2(a+b+c)(ab+bc+ca)$

Since $a^2 + b^2 + c^2 = 1$ , this leads to:

$(a+b+c)^3 = a+b+c$

Thus there are two cases to examine. Case 1: $a+b+c = 0$ . This gives one possible value of $a+b+c$ . An example of a solution of this form is $(\frac{1}{\sqrt{2}} , -\frac{1}{\sqrt{2}}, 0)$ . Case 2: $a+b+c \not= 0$ . This implies:

$(a+b+c)^2 = 1$

Which implies that $a+b+c = \pm 1$ . Examples of solutions of this form are $(1,0,0)$ and $(-1,0,0)$ . Thus $a+b+c$ can only take on 3 values.

a2+b2+c2=1 , a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=−3, (a+b+c)2=1+2(ab+bc+ca), (b+c)/a+(a+b)/c+(a+c)/b=-3, b2c+bc2+a2b+ab2+a2c+ac2=-3abc, (a+b+c)3=a+b+c+2(a2b+abc+a2c+ab2+b2c+bca+abc+c2b+c2a),(from above). (a+b+c)3=a+b+c+2(3abc+a2b+a2c+ab2+b2c+c2b+c2a), (a+b+c)3=a+b+c. so a+b+c has 3 roots

There are three possible values of $a+b+c$ : $1$ for each dimension $a$ , $b$ , and $c$ .

we can find a=b=c by solvig this. by solving we get 3 possibilities for a+b+c

Expanding the bracers and multiplying up by $3abc$ we obtain: $a^2(b+c)+b^2(a+c)+c^2(a+b)=-3abc$ Adding $a^3+b^3+c^3$ to both sides and factorising the left-hand side, we get: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2)$ -> $(a+b+c)(a^2+b^2+c^2-(ab+bc+ac))=(a+b+c)(a^2+b^2+c^2)$ Subtracting and factorising, we get: $(a+b+c)(ab+bc+ac)=0$ So either $a+b+c=0$ , which is one value of $a+b+c$ , or $ab+bc+ac=0$ -> $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$ -> $(a+b+c)^2=1$ , so $a+b+c=1$ or $a+b+c=-1$ . So there are $\boxed{3}$ possible values of $a+b+c$

First of all, we want a+b+c to show up in the second equation. Rewrite the second equation as (a+c)/b+(a+b)/c+(b+c)/a = -3. Now we can manipulate by adding b/b, c/c, and a/a to the LHS, while simply adding 3 to the RHS. This yields (a+b+c)(1/a+1/b+1/c)=0. Now either a+b+c=0, or (1/a+1/b+1/c)=0, which implies ab+bc+ca=0 if we multiply both sides by abc.

Next, we wish to make use of the first equation. We want to get squared terms, so note that (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+2(ab+bc+ca). If ab+bc+ca=0, we are left with (a+b+c)^2=1, or (a+b+c) = plus and minus 1.

So our possible values for a+b+c are: 0, +1, and -1. Thus there are 3 solutions.

I decide to try something very simple for me just take a symmetric polynomial $a+b+c$ , $ab+bc+ca$ and $abc$ but it gets to far .... so I try another thing

the next idea that came to my head is I notice the -3 on the right size of the second equation and try to think of the left size whether I could complete each parentheses as $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ so I did it . I see that the second equation will become $(a+b+c)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = 0$

so ; $a+b+c = 0$ or $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$

from the second case ... it would be $ab+bc+ca = 0$ since $a,b,c$ is not zero

from the first equation of the problem , we get $(a+b+c)^{2} - 2(ab+bc+ca) = 1$ $(a+b+c)^{2} = 1$ so $a+b+c = 1,-1$

from all of which I explain my idea ... we get that the possible value of $a+b+c = 0,1,-1$ . Therefore the number of possible values is 3

$a(\frac{1}{b}+\frac{1}{c})+b(\frac{1}{c}+\frac{1}{a})+c(\frac{1}{a}+\frac{1}{b})=-3$ $\Rightarrow a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=(a+b+c)(ab+bc+ca)=(a+b+c)((a+b+c)^2-1)/2=0$ $a+b+c=0,1,-1$