# How do you solve the system of equations: A^(A+1)+B^(B+1)+C^(C+1) = 729 and A^(1/A) = B^(1/B) = C^(1/C) ?

Apr 14, 2017

${A}^{\frac{1}{A}} \approx 1.432024273098$

#### Explanation:

The simplest conditions under which:

${A}^{\frac{1}{A}} = {B}^{\frac{1}{B}} = {C}^{\frac{1}{C}}$

are when $A = B = C$

There are other solutions, but let's stick with that for now.

In which case:

$3 \cdot {A}^{A} A = 729$

So:

${A}^{A + 1} = {A}^{A} A = \frac{729}{3} = 243 = {3}^{5}$

If the question had specified $243 = {3}^{5}$ instead of $729 = {3}^{6}$ then we would find $A = 3$ and ${A}^{\frac{1}{A}} = \sqrt{3}$. I suspect that this is what the question should have been and $729$ is in error.

As it is, the solution is not expressible in terms of elementary functions. We can find approximations using Newton's method.

Let:

$f \left(x\right) = {x}^{x + 1} - 243$

Then:

$f ' \left(x\right) = \left(x + 1\right) {x}^{x} + {x}^{x + 1} \ln x$

by Newton's method, if ${a}_{i}$ is an approximation to a zero of $f \left(x\right)$, then a better approximation is given by:

${a}_{i + 1} = {a}_{i} - \frac{f \left({a}_{i}\right)}{f ' \left({a}_{i}\right)}$

So in our example, we would probably choose ${a}_{0} = 3$ and iterate using the formula:

${a}_{i + 1} = {a}_{i} - \frac{{a}_{i}^{{a}_{i} + 1} - 243}{\left({a}_{i} + 1\right) {a}_{i}^{{a}_{i}} + {a}_{i}^{{a}_{i} + 1} \ln {a}_{i}}$

Putting this in a spreadsheet and iterating, I found:

${a}_{0} = 3$

${a}_{1} \approx 3.822386809107$

${a}_{2} \approx 3.583342155402$

${a}_{3} \approx 3.465951810855$

${a}_{4} \approx 3.443670010969$

${a}_{5} \approx 3.442996111113$

${a}_{6} \approx 3.442995518121$

${a}_{7} \approx 3.442995518121$

Using $A = 3.442995518121$, we find:

${A}^{\frac{1}{A}} \approx 1.432024273098$