Answer: [tex]\frac{8x^3}{27y^6}[/tex]
This is the fraction 8x^3 all over 27y^6
On a keyboard, we can write it as (8x^3)/(27y^6)
===========================================================
Explanation:
The exponent tells you how many copies of the base to multiply with itself.
We'll have three copies of [tex]\left(\frac{2x}{3y^2}\right)[/tex] multiplied with itself due to the cube exponent on the outside.
So,
[tex]\left(\frac{2x}{3y^2}\right)^3 = \left(\frac{2x}{3y^2}\right)*\left(\frac{2x}{3y^2}\right)*\left(\frac{2x}{3y^2}\right)\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{2x*2x*2x}{(3y^2)*(3y^2)*(3y^2)}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{(2*2*2)*(x*x*x)}{(3*3*3)*(y^2*y^2*y^2)}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{8x^3}{9y^6}\\\\[/tex]
-------------------
Or another approach you could take is to cube each component of the fraction. The rule I'm referring to is [tex]\left(\frac{a}{b}\right)^c = \frac{a^c}{b^c}[/tex]
Applying that rule will lead to:
[tex]\left(\frac{2x}{3y^2}\right)^3 = \frac{(2x)^3}{(3y^2)^3}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{2^3*x^3}{3^3*(y^2)^3}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{8x^3}{27y^{2*3}}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{8x^3}{27y^6}\\\\[/tex]
Either way you should get 8x^3 all over 27y^6 as one fraction.
