Find three perfect squares whose product is 45 less than the largest 4-digit perfect cube?

Answer:

There are quite a few solutions to this problem, and here's how to find them:

9999^1/3 = 21.543...

Hence 21³ = 9261 is the largest four digit cube number. 45 less than this is 9216.

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We want to expess 9216 in the form A²B²C²

9216 = A²B²C²

Square root both sides:

96 = ABC

And from here we can deduce possible values for A, B and C. The prime factorization of 96 is:

96 = 2⁵ x 3 = 2 x 2 x 2 x 2 x 2 x 3

By separating this string of prime factors into three groups we can find our possible solutions.

e.g. 96 = (2 x 2) x (2 x 2 x 2) x (3)

so 96 = 4 x 8 x 3

and 9216 = 4² x 8² x 3²

Find more solutions by splitting the prime factors differently.