Asked by Jackie
What is the first natural number that is a perfect square and a perfect cube? Can you find the second? The third? Explain.
Answers
Answered by
Reiny
so we want
a^2 = b^3
a = b^(3/2) or (√b)^3
so pick b = a perfect square,
e.g. b = 4 , then a = 8
so 8^2 = 4^3 = 64
let b = 9 , a = 27
9^3 = 27^2 = 729
let b = 16 , then a = 64
16^3 = 64^2 = 4096
let b = 25 , a = 125
25^3 = 125^2 = 15625
So the first few numbers with that property are:
64 , 729 . 4096 , 15625
As you can see, we can form as many of these as we want, there would be an infinite number of them.
a^2 = b^3
a = b^(3/2) or (√b)^3
so pick b = a perfect square,
e.g. b = 4 , then a = 8
so 8^2 = 4^3 = 64
let b = 9 , a = 27
9^3 = 27^2 = 729
let b = 16 , then a = 64
16^3 = 64^2 = 4096
let b = 25 , a = 125
25^3 = 125^2 = 15625
So the first few numbers with that property are:
64 , 729 . 4096 , 15625
As you can see, we can form as many of these as we want, there would be an infinite number of them.
Answered by
Jackie
Ok thank you. I understand now.
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