# Which number in the monomial 125x^18y^3z^25 needs to be changed to make it a perfect cube?

Oct 19, 2016

$z$'s exponent would need to be changed from $25$ to a multiple of $3$

#### Explanation:

Using the properties ${\left({x}^{a}\right)}^{b} = {x}^{a b}$ and ${\left(x y\right)}^{a} = {x}^{a} {y}^{a}$, we can see that ${\left({x}^{a} {y}^{b}\right)}^{3} = {\left({x}^{a}\right)}^{3} {\left({y}^{b}\right)}^{3} = {x}^{3 a} {y}^{3 b}$.

Notice that the $3$ gets distributed to each of the exponents, meaning each of the exponents in the separated form has to be a multiple of $3$. Reversing these steps shows that a product of integers and variables is a perfect cube only if their powers are all multiples of $3$.

In this case:
$125 = {5}^{3 \cdot 1}$
${x}^{18} = {x}^{3 \cdot 6}$
${y}^{3} = {y}^{3 \cdot 1}$
${z}^{25} = {z}^{3 \cdot 8 + \textcolor{red}{1}}$

Notice that the exponent of $z$ is not a multiple of $3$, meaning that would need to be changed in order for the product to be a perfect cube. If it was, say, $24 = 3 \cdot 8$, then we could write the product as

$125 {x}^{18} {y}^{3} {z}^{24} = {5}^{3} {x}^{6 \cdot 3} {y}^{3} {z}^{8 \cdot 3}$

$= {5}^{3} {\left({x}^{6}\right)}^{3} {y}^{3} {\left({z}^{8}\right)}^{3}$

$= {\left(5 {x}^{6} y {z}^{8}\right)}^{3}$

which is a perfect cube.