Let a + ar + ar^2 + ar^3 + \dotsb be an infinite geometric series. The sum of the series is $4.$ The sum of the cubes of all the terms is $10.$ Find the common ratio.
\(a+ar+ar^2+...=\frac{a}{1-r}\) as long as r<1
\(a^3+a^3(r^3)+a^3(r^3)^2+...=\frac{a^3}{1-r^3}\) as long as r^3<1
So:
\(\frac{a}{1-r}=4\)
and
\(\frac{a^3}{1-r^3}=10\)
Can you take it from here?
+1427 
