Algebra

The sum of the infinite geometric series listed is \(\frac{a}{1-r}\), and this equals four, so \(\frac{a}{1-r}=4\).

The series made up of all the cubes of these terms is \({a}^3+{a}^{3}{r}^{3}+{a}^3{r}^{6}+{a}^{3}{r}^{9}...\). Because \(r<1\), then \({r}^{3}<1\). So applying our formula again, \(\frac{{a}^3}{1-{r}^3}=10\). Now we have \(a=4(1-r)\), and \({a}^{3}=10(1-{r}^{3})\). We can substitute the first equation in the second one, \(64{(1-r)}^{3}=10(1-{r}^{3})\). 

\(64(1-3r+3{r}^{2}-{r}^{3})=10-10{r}^{3}\)

\(54-192r+192{r}^{2}-54{r}^{3}=0\).

\(9-32r+32{r}^{2}-9{r}^{3}=0\)

This is factored into

\(-(r-1)(9{r}^{2}-23r+9)=0\).

We know r is not 1, so

\(9{r}^2-23r+9=0\).

We solve and discard one solution, getting​\(\frac{23-\sqrt{205}}{18}\)as the common ratio.

\(4 = a + ar + ar^2 + ar^3 + \dotsb\)

\(10 = a^3 + a^3r^3 + a^3r^6 + \dotsb\)  Find \(r\)

Infinite geometric series formula: a1 [first term] divided by (1 - r) [1 - common ratio] = \(a_1\over{(1 - r)}\)

Using this piece of information, we can rewrite the equations from earlier:

\(4 = {a\over{(1 - r)}}\); \(4 - 4r = a\)

\(10 = {a^3\over{1 - r^3}}\); \(10 - 10r^3=a^3=(4-4r)^3=-64r^3+192r^2-192r+64\)

\(10 - 10r^3 = -64r^3+192r^2-192r+64\); \(54r^3-192r^2+192r-54=0\); \(54r^3-54 - 192r^2+192r=0\)

Factor by grouping: This part is hard...

\(54(r^3-1)-192r(r-1)=0\)

\(r^3-1=(r-1)(r^2+r+1)\); remember this!!! Factoring difference/sum of cubes is important!!

\(54(r-1)(r^2+r+1)-192(r-1)=0\); \((r - 1)(54r^2+54r-138)=0\)

We know the common ratio cannot be 1, because then the series would not converge.

\(54r^2+54r-138=0\); divide by 54: \(r^2+r-{23\over9}=0\)

By quadratic formula:

\(r = {-1 \pm \sqrt{1+{92\over9}} \over 2}={-1\pm{\sqrt{101}\over3}\over2}=-{1\over2}\pm{\sqrt{101}\over6}\)

please reply if you catch any errors... because this number is ugly :/