Answered

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without using a calculator prove whether 1728 is a perfect cube

Sagot :

[tex]^3\sqrt[]{1728}^{}^{}^{}[/tex][tex]^3\sqrt[]{(2\cdot2\cdot2)(2\cdot2\cdot2)(3\cdot3\cdot3)}[/tex]

Since the prime factors of 1728 can be grouped into triples of equal factors, it is a perfect cube.

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