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 :/