Which of these equations has a rational solution?
A: 2(over)5 x^3=26
B: 3(over)5 x^2=54
C: 2(over)5 x^3=50
Can you please explain each stage to achieving the answer, I'm really stuck and don't understand how to answer this!
A: 2(over)5 x^3=26
B: 3(over)5 x^2=54
C: 2(over)5 x^3=50
Can you please explain each stage to achieving the answer, I'm really stuck and don't understand how to answer this!

A: 2over 5 x^3=26
2x^3=26*5
2x^3=130
x^3=130/2
x^3=65
x=cube root of 65
irrationnal solution
B: 3over 5 x^3=54
3x^3=54*5
3x^3=270
x^3=270/3
x=cube root of 90
irrationnal
C: 2over 5 x^3=50
2x^3=50*5
2x^3=250
x^3=250/2
x^3=125
x=cube root of 125
x=5, rationnal
2x^3=26*5
2x^3=130
x^3=130/2
x^3=65
x=cube root of 65
irrationnal solution
B: 3over 5 x^3=54
3x^3=54*5
3x^3=270
x^3=270/3
x=cube root of 90
irrationnal
C: 2over 5 x^3=50
2x^3=50*5
2x^3=250
x^3=250/2
x^3=125
x=cube root of 125
x=5, rationnal

rearrange them to get them so that they show x to a power equals a number like so:
x^3=26*5/2 = 65
x^2=54*5/3 = 90
x^3 = 50*5/2 = 125
rational solutions to these will only occur when the number on the right hand side is a perfect square or cube (depending on the power on the left hand side). You need to be able to recognise your perfect squares and perfect cubes.
65 is not a cubed number
90 is not a squared number
125 is the cube of 5, so the answer is C
x^3=26*5/2 = 65
x^2=54*5/3 = 90
x^3 = 50*5/2 = 125
rational solutions to these will only occur when the number on the right hand side is a perfect square or cube (depending on the power on the left hand side). You need to be able to recognise your perfect squares and perfect cubes.
65 is not a cubed number
90 is not a squared number
125 is the cube of 5, so the answer is C