# What is the cube root of 128?

Feb 4, 2015

By definition, the cubic root of a number $x$ is a number $y$ such that ${y}^{3} = x$.

Apart from using the calculator, of course, you can see if a number $n$ is a perfect square by factoring it into primes, and if the number has a representation of the form
$n = {p}_{1}^{{d}_{1}} \setminus \times {p}_{2}^{{d}_{2}} \setminus \times \ldots \setminus \times {p}_{n}^{{d}_{n}}$, then it is a perfect cube if and only if every ${d}_{i}$ is divisible by 3.

Factoring $128$ in primes gives you
$128 = {2}^{7}$, thus it is not a perfect cube (i.e., its cube root is not an integer).

Anyway, we can say that the cubic root of $128$ is $128$ to the power of $\frac{1}{3}$, so we have
${128}^{\frac{1}{3}} = {\left({2}^{7}\right)}^{\frac{1}{3}} = {2}^{\frac{7}{3}} = {2}^{2 + \frac{1}{3}}$

By using the formula ${a}^{b + c} = {a}^{b} \setminus \cdot {a}^{c}$, we have that
${2}^{2 + \frac{1}{3}} = {2}^{2} \setminus \cdot {2}^{\frac{1}{3}} = 4 \setminus \cdot {2}^{\frac{1}{3}}$
which is four times the cubic root of $2$