Check this equation for me please

No, this is VERY easy, you don't need to be smart to solve this -.- you have:
⅓π (x/3)²h = 3π*x*x/3
the pi's cancel with each other cause there is one on each side:
⅓ (x/3)²h = 3*x*x/3, then do a little math on both side you get:
1/3 * (x^2/9)*h = x^2 which equals to: (1/3 *1/9 = 1/27)
((x^2)/27)*h = x^2 Cancel the x^2 on each side,
h/27 = 1 cross multiply
h = 27

⅓π (x/3)²h = 3π*x*x/3
Make h the subject by dividing 3π*x*x/3 by ⅓π (x/3)²

h = [3π*x*x/3] / [⅓π (x/3)²]
*** numerator and denominator both have 'x/3' so cancel one on each side. ***
= [3π*x] / [⅓π (x/3)] multiply both sides by 3 to get rid of the ⅓ on the denominator.
= [9π*x] / [π (x/3)] multiply both sides by '3' to make the 'x/3' become an 'x'.
= 27πx / πx then cancel out 'πx' on both sides.
= 27
∴ h = 27

Now I remember why I dropped out!! Good luck kid!! If you know how to solve those equations, your a smart lad