Help me with this this system of equations

Question

Help me with this system of equations $$a+b = 3 -c$$ $$\frac{1}{a}+\frac{1}{b}= \frac{5}{12}-\frac{1}{c}$$ $$ a^3+b^3 = 45 -c^3$$

Answer 1

We have

$$\begin{align} a+b+c & = 3 \\ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} &= \frac{5}{12} \\ a^3 + b^3 + c^3 & = 45 \\ \end{align} $$

The second equation tells us that $\frac{ab+bc+ca}{abc} = \frac{5}{12} $, or that $12(ab+bc+ca) = 5(abc)$.

Using $(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a) = 3(3-a)(3-b)(3-c)$, we get that $27-45 = 3[27-9(a+b+c) + 3(ab+bc+ca) - abc]$, or that

$$-6 = 3(ab+bc+ca) - abc.$$

Hence, we get that $ab+bc+ca = -10 $, $abc = -24 $. by solving the simultaneous equations.

Hence, using Vieta, we get that $a, b, c$ are (permutation of) roots to the cubic

$$x^3 - 3x^2 - 10 x + 24 = 0. $$

This cubic is $(x-2)(x-4)(x+3) = 0$, and thus we have $\{a,b,c\} = \{2, 4, -3\} $.

Answer 2

First of all notice that $a$ and $b$ appear "symmetrically": if you exchange $a$ with $b$, the system doesn't change. In these cases, setting

$$x=a+b,\quad y=ab$$

usually helps.

Indeed, you can note now that the left hand side of the second equation is just $x/y$, while

$$ a^3+b^3 = x^3 -3xy $$

The first equation already tells you what $x$ is; so the third equation tells you what $y$ is. Hence the second equation will give you the value (or values) of $c$. Hence you reduce your problem to finding the numbers $a$ and $b$ knowing their sum and product.

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A complete solution

We can write $$ \begin{cases} x=3-c\\[2ex] \frac{x}{y}=\frac{5c-12}{12c}\\[1ex] x^3-3xy=45-c^3 \end{cases} $$ The third equation becomes $$(3-c)^3-3(3-c)y=45-c^3$$ or $$27-27c+9c^2-c^3-3(3-c)y=45-c^3$$ from which, rearranging terms and simplifying, $$(3-c)y=3c^2-9c-6$$ or $$y=\frac{3c^2-9c-6}{3-c}$$ Thus $$\frac{x}{y}=\frac{(3-c)^2}{3(c^2-3c-2)}$$ and the second equation becomes $$4c(3-c)^2=(c^2-3c-2)(5c-12).$$ It's a matter of computing, now: $$36c-24c^2+4c^3=5c^3-15c^2-10c-12c^2+36c+24$$ $$c^3-3c^2-10c+24=0$$ It's easy to see that this has the roots $c=-3$, $c=2$ and $c=4$. Let's see what you get from $c=-3$: $$ \begin{cases} a+b=6\\[1ex] \frac{a+b}{ab}=\frac{3}{4} \end{cases} $$ or $$\begin{cases} a+b=6\\[1ex] ab=8 \end{cases} $$ that can be solved through the auxiliary equation $t^2-6t+8=0$. This one has solutions $t=2$ and $t=4$. Thus you get the solutions $$ \begin{cases}a=2\\ b=4\\ c=-3 \end{cases} \qquad \begin{cases} a=2\\ b=4\\ c=-3 \end{cases} $$ Observing that the system is actually symmetric in $a$, $b$ and $c$, the other solutions are obtained by permuting the first one.