RATIONAL a and b such that a^3=b^2?

Question

Besides a and b both being 0 or 1, is there any other rational numbers a and b such that a^3=b^2?

i obviously missed the obvious. Thanks guys

Answer 1

Consider any 6th power, n^6, this can be written as either (a^3)^2, or (b^2)^3.

So for instance you have 4^3 = 8^2 = 64; 9^3 = 27^2 = 729 and so on.

Answer 2

Going along with Swamy's answer, you can create even more numbers on the list if you include fractions. For example consider 1/2^6. a = 1/4 and b = 1/8. Another ex.: (3/7)^6, then a = 9/49, b = 27/343.
You can work with any rational raised to the sixth power (that is, they needn't be whole numbers).

Answer 3

a^3=b^2=n^6 (let)
this implies a=n^2 and b=n^3
n a b a^3=b^2=n^6
0 0 0 0
1 1 1 1
2 4 8 64
3
4
5
......
fill up the list. it goes.
0,1,64,3^6,4^6,5^6,.......
As slaga added to my answer, the value of n need not be natural numbers. n can take any real value.
This technique is true not only for squres and cubes but also true for any two powers say m and n.
we find the values of a and b such that a^m=b^n taking it as equal to some p^(mn), and a=p^n & b=p^m.

Answer 4

If you substitute c^6 for a, you'll find that many rational numbers fit the equation. For example, if c=2, then c^6=64, (c^6)^3=2^18=b^2, b=2^9, a rational number.

Answer 5

a=4 b=8 ?

Answer 6

alrighty then

aaa=bb

since you know that a
you have to think of perfect squares

4 16 25 36 49 64 81

now take each variable and see if you can square it or cube it to get one of those numbers

the cube root of 64 is 4 and the squareroot of 64 is 8

so 8=8

now you have to square each variable (meaning only 1 a and one b so 4 and 8)
and you get

16^3=64^2
4096=4096

to get continuous results you square each variable (a and b) and the equation will always come out equal
16^2=256
64^2=4096
256^3=4096^2

16777216=16777216

Answer 7

Lots.

Any for which b = sqrt (a ^ 3)

eg a = 4, b = 8

Answer 8

a= 100
b = 1000

100*100*100=1000 *1000