Question

. Is 5625000 a perfect cube? Explain.

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,

Let's look into the case , a really interesting question!

Now , the given number or value , is ; "5625000"

By the methodology of prime factorisation (can also work by Euclid's Lemma) ,, here ,,

$= > 5625000 = > 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5$
By grouping the given grouping of three prime factors or prime numbers ,, we get ,,

$= > {(2)}^{3} \times {(5)}^{3} \times {(5)}^{3} \\ \\ = > 2 \times 5 \times 5 \\ \\ = > 50$
Now ,, if we see there are more than one extra prime factor or number , so , more than should be taken into consideration for the total value to obtained , finally , as "5625000" .

But , again if we properly observe the prime factors and other three number not fitting the criteria of Grouping (in three pairs) , "3" , "3" and "5" . Now taking the cube root of the following , that is ,

$= > \sqrt[3]{ {2}^{3} \times {5}^{3} \times {5}^{3} } \\ \\ = > \sqrt[3]{8 \times 125 \times 125} \\ \\ = > \sqrt[ 3]{125000} \\ \\ = > 50$
The numbers which were left ,,,

$= > 50 \times \sqrt[3]{3 \times 3 \times 5} \\ \\ = > 50 \times \sqrt[3]{45} \\ \\ = > 50 \times 3.5569 \\ \\ = > 177.8447$
By multiplying it (the value 177.8447) three times,,,,

$= > {177.84467}^{3} \\ \\ = > 5624999.999 \\ \\ = > 5625000$
The value which is making the imperfect cube is ,,,

$= > \sqrt[3]{45} \\ \\ or \\ \\ = > 3.5569$
So this disables the value of "5625000" to verify as a perfect cube .

OR The value of 5625000 is a $\textbf{Imperfect Cube}$

HOPE IT HELPS YOU AND GIVES A INSIGHT OF IMPERFECT CUBES!!!!!!