Well if you have three and add negative 3 it equals 0
J ---- P --- K
JP = 2x
PK = 7x
JK = 27
2x + 7x = 27
9x = 27
x = 27/9
x = 3
The value of P is 3.
JP = 2x = 2(3) = 6
PK = 7x = 7(3) = 21
Rational exponents

work like this: the numerator is the actual exponent of the base, while the denominator is the index of the root.
In other words, we have
![a^\frac{m}{n}=\sqrt[n]{a^m}](https://tex.z-dn.net/?f=a%5E%5Cfrac%7Bm%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Ba%5Em%7D)
So, in you case, we have
![\sqrt[5]{x^3}=x^\frac{3}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E3%7D%3Dx%5E%5Cfrac%7B3%7D%7B5%7D)
Assuming that the question contanis a typo. If you actually mean
,
then you can write it as 