Q:

Time complexity of $\sim c \cdot n^3$ with a computer which is 10 times faster

I am trying to solve the following question

There's an algorithm with time complexity $\sim c \cdot n^3$. Suppose
  there's another computer which is 10 times faster. How much bigger can
  our $n$ be to be solved in the same time?

My answer so far is $\sqrt[3]{10}$.
$n^3$ can be 10 times bigger, so I have to multiply N with $\sqrt[3]{10}$. Is my conclusion correct or is there a fault in my reasoning?

A:

You've correctly reasoned that the additional speed doesn't change the $n^3$ part of the tilde expression. The only component left for it to effect then is $c$. 
Let $c_1$ be the constant of the first computer and $c_2$ the constant of the second. If the second computer is 10x faster, then $c_2 = c_1/10.$ Now, all you need to do is solve the following equation:
$$c_1 \cdot n^3 = \frac{c_1 \cdot n'^3}{10},$$
which gives the same answer you came up with!

