The 3 (Not) Real Values

This can easily be solved backwards: first we manipulate the last equation a bit until it reads $31=\frac{a^2+b^2+c^2+3(ab+bc+ac)}{(a+b)(b+c)(a+c)} = \frac{23+3\large{(}\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}\large{)}}{(a+b)(b+c)(a+c)}=\frac{23+\frac{3}{2}(49-23)}{(a+b)(b+c)(a+c)},$ which equals $\frac{62}{(a+b)(b+c)(a+c)}.$

The next realisation is that $3(a+b)(b+c)(a+c)=(a+b+c)^3-(a^3+b^3+c^3) = 7^3 -(a^3+b^3+c^3).$

Combining these two equations, we find that $a^3+b^3+c^3=(a+b+c)^3 - 3(a+b)(b+c)(a+c) = 7^3 - 3\cdot\frac{62}{31}=7^3-6 = 337.$

$\dfrac{1}{a+b} + \dfrac{1}{b+c} + \dfrac{1}{c+a} = 31$ . Simplification leads to

$\dfrac{a^2 + b^2+c^2+3ab+3bc+3ac}{(a+b)(b+c)(c+a)} = 31$ .

Now, $ab + bc+ac=\dfrac{(a+b+c)^2 -(a^2 + b^2 + c^2)}{2} = 13$ . So we have,

$\dfrac{23+39}{(a+b)(b+c)(c+a)} = 31$ $\implies$ $(a+b)(b+c)(c+a) = 2$

Finally $\large{a^3+b^3+c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a) = 7^3 - 3 \times 2 = 343 - 6 = \boxed{337}}$

$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2*(ab+bc+ac)$ Hence $7^{2} = 49 = 23 + 2*(ab+bc+ac)$ Therefore (ab+bc+ac) = 13 -------> (1)

$(a+b+c)^{3} = a^{3}+b^{3}+c^{3}+3*(a+b)(b+c)(c+a)\[ Hence \[7^{3} = 343 = a^{3}+b^{3}+c^{3} +3*(a+b)(b+c)(c+a)$ Therefore $a^{3}+b^{3}+c^{3} = 343 - 3*(a+b)(b+c)(c+a)$ -----------> (2)

We are given that $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} = 31$ which is on simplification $\frac{a^{2}+b^{2}+c^{2}+3*(ab+bc+ac)}{(a+b)(b+c)*(c+a)} = 31$ Which is $\frac{23+3*13}{(a+b)(b+c)(c+a)} = 31$ Hence (a+b)(b+c)*(c+a) = 62/31 = 2 -------------------> (3)

Substituting (3) in (2) we get $a^{3}+b^{3}+c^{3} = 343 - 3*2 = \boxed{337}$

We'll use this identity to find the value of $a^3 + b^3 + c^3$ : $(a+b+c)^3=a^3+b^3+c^3 +3(a+b+c)(ab+bc+ac)-3(abc)$

From 1st and 2nd equation in the problem, we can find $ab+bc+ac$ $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ac)$

$7^2= 23 + 2(ab+bc+ac)$

$ab+bc+ac= 13$

Meanwhile, we can write the 3rd equation into this form : $\frac{1}{7-a}+\frac{1}{7-b}+\frac{1}{7-c}=31$ .

Then, $\frac{(7-a)(7-b)+(7-b)(7-c)+(7-c)(7-a)}{(7-a)(7-b)(7-c)}=31$

$(7-a)(7-b)+(7-b)(7-c)+(7-c)(7-a)=31(7-a)(7-b)(7-c)$

$3.7^2 -14(a+b+c)-(ab+bc+ac) = 31(7^3-7^2(a+b+c)+7(ab+bc+ac)-abc)$

$3.7^2 - 14(7) - (13) = 31 (7^3-7^3+7(13)-abc)$

$abc =89$

Substitute $ab+bc+ac= 13, abc =89$ and $a+b+c=7$ into $(a+b+c)^3=a^3+b^3+c^3 +3(a+b+c)(ab+bc+ac)-3(abc)$ .

Then, $(a+b+c)^3=a^3+b^3+c^3 +3(a+b+c)(ab+bc+ac)-3(abc)$

$(7)^3=a^3+b^3+c^3 +3(7)(13)-3(89)$

$a^3+b^3+c^3 = \boxed{337}$

I have computed the answer to be $337$ ; however, the truth is that no such real numbers exist!!

Using the third equation $\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{a + c} = 31$ , we can show that $abc = 89$ (Detailed proof at the end of this post.) Then you will recall that the AM-GM inequality holds for all nonnegative real numbers, so in particular consider $a^2, b^2,$ and $c^2$ . We have that $\begin{aligned} \frac{a^2 + b^2 + c^2}{3} &\ge \sqrt[3]{a^2 b^2 c^2} \\ \frac{23}{3} &\ge \sqrt[3]{(abc)^2} \\ \frac{23}{3} &\ge \sqrt[3]{89^2} \\ \frac{23^3}{3^3} &\ge 89^2 \\ \frac{12167}{27} &\ge 7921 \\ 12167 &\ge 7921 \cdot 27 = 213867 \end{aligned}$ which is a contradiction, showing that $a,b,c$ do not, in fact, exist.

Another way to see that no real numbers exist is to compute the polynomial $(x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + bc + ca)x - abc$ . It is straightforward to show that $ab + bc + ca = 13$ (this is at the end of this post also), so the polynomial is $x^3 - 7x^2 + 13x - 89$ which one can check has no real roots, either by calculating that the discriminant is negative , or by graphing the equation .

Over the complex numbers these sorts of problems are guaranteed a solution, but over the real numbers they are not. One should be careful when formulating such a problem to make sure there is actually a solution!

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Here are proofs of the basic facts used above.

First, $ab + bc + ca = \frac{(a + b + c)^2 - (a^2 + b^2 + c^2)}{2} = \frac{7^2 - 23}{2} = 13$ Second, $\begin{aligned} \frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} &= 31 \\ \implies (a + c)(b + c) + (a + b)(a + c) + (a + b)(b + c) &= 31(a + b)(b + c)(c + a) \\ \end{aligned}$ So we have $\begin{aligned} a^2 + b^2 + c^2 + 3(ab + bc + ca) &= 31[a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc] \\ a^2 + b^2 + c^2 + 3(ab + bc + ca) &= 31[(a + b + c)(ab + bc + ca) - abc] \\ a^2 + b^2 + c^2 + 3(ab + bc + ca) &= 31(a + b + c)(ab + bc + ca) - 31abc \end{aligned}$ Therefore, $\begin{aligned} 31abc &= 31(a + b + c)(ab + bc + ca) - 3(ab + bc + ca) - (a^2 + b^2 + c^2) \\ &= 31(7)(13) - 3(13) - 23 = 31(91) - 39 - 23 = 31(91) - 31(2) \\ \implies abc &= 91 - 2 = 89. \end{aligned}$

First, note the following general equations for values $a$ , $b$ , and $c$ :

1. $(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$
2. $(a+b+c)^3=a^3+b^3+c^3+3(a^2(b+c)+b^2(a+c)+c^2(a+b)+2abc)$
3. $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}=\frac{a^2+b^2+c^2+3(ab+ac+bc)}{(a+b)(b+c)(a+c)}$
4. $(a+b)(b+c)(a+c)=a^2(b+c)+b^2(a+c)+c^2(a+b)+2abc$

Here, we can see that:

• Using equation $1$ , the value of $ab+ac+bc$ is $\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=\frac{49-23}{2}=13$ .
• Letting the right hand side of equation $3$ be equal to the given value of $31$ and multiplying the denominator over, we get that $a^2+b^2+c^2+3(ab+ac+bc)=23+3(13)=62=31(a+b)(b+c)(a+c)$ $\Rightarrow (a+b)(b+c)(a+c)=2$ .
• Note that, using equation $4$ , we can make a substitution and rewrite equation $2$ : $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$ . Thus, solving for $a^3+b^3+c^3$ yields $a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(a+c)=343-3(2)=\boxed{337}$ .

a + b + c = 7

$a^2 + b^2 + c^2$ = 23

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

$7^2$ = 23 + 2( ab + bc + ca )

49 = 23 + 2( ab + bc + ca )

2( ab + bc + ca ) = 49 - 23

( ab + bc + ca ) = 13

( ab + bc + ca )( a + b + c ) = 91

(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) + 3abc = 91

Simplify : $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}$ .

$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}$ = $\frac{3(ab + bc + ac) + (a^2 + b^2 + c^2)}{2abc + (a^2b + a^2c + b^2a + b^2c + c^2a + c^2b)}$

$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c}$ = 31

$\frac{3(ab + bc + ac) + (a^2 + b^2 + c^2)}{2abc + (a^2b + a^2c + b^2a + b^2c + c^2a + c^2b)}$ = 31

31 = $\frac{3(ab + bc + ac) + (a^2 + b^2 + c^2)}{3abc + (a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) - abc }$

[adding and subtracting 'abc' to the denominator]

31 = $\frac{3(13) + (23)}{3abc + (a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) - abc }$

$\frac{39 + 23}{91 - abc }$ = 31

$\frac{62}{91 - abc }$ = 31

91 - abc = 2

abc = 89

$a^3 + b^3 + c^3$ - 3abc = (a + b + c)[ $a^2 + b^2 + c^2$ - (ab + bc + ac)]

$a^3 + b^3 + c^3$ - 3(89) = 7(23 - 13)

$a^3 + b^3 + c^3$ - 267 = 7(10)

$a^3 + b^3 + c^3$ = 267 + 70

$a^3 + b^3 + c^3$ = 337

(a + b + c)3= a^3 + b^3+ c^3 + 3 [(a+b)(b+c)(a+c)] ----1

given

31 = [a^2+b^2+c^2 + 3(ab+bc+ac)] / [(a+b)(b+c)(a+c)]

== > [ (a+b)(b+c)(a+c)] = [a^2+b^2+c^2 + {3/2[(a + b + c)^2 - a^2+b^2+c^2]}] / 31

                                       =   [23 + { 3/2  [(7)^2 - 23]}]  / 31

                                       = [23 + { 3/2 [26] } ]  /  31

                                         =[23+ 39 ]   /  31

                                          =62 / 31  = 2 =    [ (a+b)(b+c)(a+c)]------2

                            substitute 2 in 1

                     7^3  =    a^3 + b^3+ c^3+  3(2)
                   343 - 6  =  a^3 + b^3+ c^3
                 ans:-  a^3 + b^3+ c^3  =  337

(a+b+c)^3=a3+b3+c3+3(a2b+ab2+b2c+bc2+a2c+ac2)+6abc ab+bc+ca=((a+b+c)^2-(a2+b2+c2))/2=13 (a+b)(b+c)(c+a)=a2b+ab2+b2c+bc2+a2c+ac2+2abc=2 a2b+ab2+b2c+bc2+a2c+ac2=2-2abc now put all these values in first eqn. (a+b+c)^3=a3+b3+c3+3(a2b+ab2+b2c+bc2+a2c+ac2)+6abc 343 = a3+b3+c3+3(2-2abc)+6abc 343-6=a3+b3+c3 therefore a3+b3+c3=337

Step1: get a^{2}b + b^{2}c + c^{2}a + ab^{2} + bc^{2} + ac^{2} + 2abc = 2 Step 2: Expand (a + b + c)^{3} and get the answer

from (a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2 (ab+bc+ca) , find ab+bc+ca; then from 3rd equation, by cross multiplication find (2 abc+a^{2}b+a b^{2}+....); then use (a+b+c)^{3} = a^{3}+b^{3}+c^{3}+3 (2 abc+a^{2}b+a b^{2}+....) to find answer....

These 3 equations can be reduced into linear equation with two variables such that 1) y=3x+70 2) y=2x+159 . where y= $\sum { a^{ 3 } }$ and x = $abc$ .

$2(ab+bc+ca)=(a+b+c)^{2}-(a^{2}+b^{2}+c^{2})=49-23=26$

or, $ab+bc+ca=13$

From the last equation, $\sum_{cyc}(a+b)(b+c)=31(a+b)(b+c)(c+a)$

or, $\sum_{cyc}(ab+ac+b^{2}+bc)=31(a+b)(b+c)(c+a)$

or, $\sum_{cyc}(ab+bc+ca)+\sum_{cyc}b^{2}=31(a+b)(b+c)(c+a)$

or, $31(a+b)(b+c)(c+a)=3.13+23=62$ .Hence, $(a+b)(b+c)(c+a)=2$

Now, $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc=91-abc$ .

Hence, $abc=89$

Now, $a,b,c$ are the roots of $x^{3}-7x^{2}+13x-89=0$

Finally, i use the result that if $a,b,c$ are the roots of $x^{3}+p_{1}x^{2}+p_{2}x+p_{3}=0$ ,then $a^{3},b^{3},c^{3}$ are the roots of $x^{3}-(p_{1}^{3}-3p_{1}p_{2}+3p_{3})x^{2}$ $+(p_{2}^{3}-3p_{1}p_{2}p_{3}+3p_{3}^{2})x$ $+$ $p_{3}^{3}=0$

Here, we just have to compute the second co-efficient.Hence the value of $a^{3}+b^{3}+c^{3}=-((-7)^{3}-3(-7)(13)+3(-89))=337$

$a+b+c=7 <=> (a+b+c)^{2} = 49 = a^{2}+b^{2}+c^{2}+2ab+2bc+2ac$

$<=>ab+bc+ac = \frac{1}{2}(49-(a^{2}+b^{2}+c^{2}))=\frac{1}{2}(49-23)=13$

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

$= \frac{a^{2}+b^{2}+c^{2}+3ab+3bc+3ac}{(a+b)(b+c)(a+c)}=\frac{(a+b+c)^2+ab+bc+ac}{(a+b)(b+c)(a+c)}$

$<=> (a+b)(b+c)(a+c) = \frac{(a+b+c)^2+ab+bc+ac}{31} = \frac{49+13}{31} = 2$

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

$<=> a^{3}+b^{3}+c^{3} = (a+b+c)^{3} - 3(a+b)(b+c)(a+c) = 7^{3} - 3*2 = 337$

(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a).............1 from the last equation we get 31(a+b)(b+c)(c+a)=(b+c)(c+a)+(a+b)(b+c)+(a+b)(c+a) =a^2+b^2+c^2+3(ab+bc+ca).................2 from the given first equation squaring both sides we get (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca) 49=23+2(ab+bc+ca) ab=bc+ca=13 substitute in ....2 we get 31(a+b)(b+c)(c+a)=23+29=62 (a+b)(b+c)(c+a)=2 substitute in ....1 we get a^3+b^3+c^3=343-6=337

$a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)$ $a^{3}+b^{3}+c^{3} = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ac)+3abc$ $a^{3}+b^{3}+c^{3} = 7(23-ab-bc-ac)+3abc$ $\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} = 31$ $\frac{(b+c)(a+c)+(a+b)(a+c)+(a+b)(b+c)}{(a+b)(b+c)(a+c)} = 31$ $\frac{ab+bc+ac+c^{2}+a^{2}+ac+ab+bc+ab+ac+b^{2}+bc}{(a+b)(b+c)(a+c)} = 31$ $\frac{a^{2}+b^{2}+c^{2}+3(ab+bc+ac)}{(a+b)(b+c)(a+c)} = 31$ $\frac{23+3(ab+bc+ac)}{(a+b)(b+c)(a+c)} = 31$ $a+b = 7-c$ $a+c = 7-b$ $b+c = 7-a$ $\frac{(b+c)(a+c)+(a+b)(a+c)+(a+b)(b+c)}{(a+b)(b+c)(a+c)} = 31$ $\frac{(7-a)(7-b)+(7-c)(7-b)+(7-c)(7-a)}{(a+b)(b+c)(a+c)} = 31$ $\frac{49-7b-7a+ab+49-7b-7c+bc+49-7a-7c+ac}{(a+b)(b+c)(a+c)} = 31$ $\frac{147-14a-14b-14c+ab+bc+ac}{(a+b)(b+c)(a+c)} = 31$ $\frac{147-14(a+b+c)+ab+bc+ac}{(a+b)(b+c)(a+c)} = 31$ $\frac{147-14(7)+ab+bc+ac}{(a+b)(b+c)(a+c)} = 31$ $\frac{147-98+ab+bc+ac}{(a+b)(b+c)(a+c)} = 31$ $\frac{49+ab+bc+ac}{(a+b)(b+c)(a+c)} = 31$ $23+3(ab+bc+ac) = 49+ab+bc+ac$ $2(ab+bc+ac) = 49-23 = 26$ $ab+bc+ac = 13$ $49+13 = 62$ $(a+b)(b+c)(a+c)=2$ $a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)+2abc=2$ $a^{2}(7-a)+b^{2}(7-b)+c^{2}(7-c)+2abc=2$ $7a^{2}-a^{3}+7b^{2}-b^{3}+7c^{2}-c^{3}+2abc=2$ $7(a^{2}+b^{2}+c^{2})-7(23-ab-bc-ac)-3abc+2abc=2$ $7(23)-7(10)-abc=2$ $161-70-2=abc$ $abc=89$ $3abc=267$ $267+70=\boxed{337}$

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

1. (a+b+c)^2 =a ^ 2 +b ^ 2 + c ^ 2 +2*(ab +bc+ca)

solving 1 we get ab+bc+ca=13

1. 1/(b+c) +1/(c+a) + 1/(b+a)=31

== a^2+b^2+c^2 +3(ab+bc+ca)/(a+b)(b+c)(c+a)=31

or (a+b)(b+c)(c+a)=2

or ba^2 + ab^2 +ca^2 +ac^2 + bc^2 + cb^2 + 2abc=2

adding abc both side

we get lhs=rhs

(a+b+c)(bc+ca+ab)=2+abc

or abc=89

put all the values to get a^3 + b^3 +c^3 = 337

From the 1st and 2nd equations we get ab + bc + ca = 13; Plug the values of a + b+ c, ab + bc + ca in the 3rd equation to obtain abc = 89. So a, b, c are the roots of the equation x^3 -7x^2 + 13x - 89 = 0. Hence a^3 + b^3 + c^3 = 7(a^2 + b^2 +c^2) - 13(a + b + c) +3(89) = 337