Math, asked by yash2090, 1 year ago

find the cube root of the following prime factorization


(i) 5832
(ii) 1728
(iii) 216000
(iv) 21952



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rahul077: 9261
allison2134: wrong
yash2090: what is 9261 wrong
rahul077: If its prime factorization can be separated into three equal groupings, the number is a perfect cube: ... Therefore, 9261 is a perfect cube withcube root=3×7=21. The cube root of 9261 is 21. Once again, short division was used to break down 9261
rahul077: correct option is not given. yash2090
rahul077: it'a right answer bro 9261
yash2090: but I didn't ask the cube root of 9261
rahul077: 18 12 60 28

Answers

Answered by Anonymous
11

(¡)5832

=>2×2×2×3×3×3×3×3×3

=> 18 ×18×18

 =  >  {(18)}^{3}

(¡¡) 1728

=>2×2×2×2×2×2×3×3×3

=> 12 ×12×12

 =  >  {(12)}^{3}

(¡¡¡) 216000

=>2×2×2×2×2×2×3×3×3×5×5×5

=> 60×60×60

 =  >  {(60)}^{3}

(iv) 21952

=>2×2×2×2×2×2×7×7×7

=> 28×28×28

 =  >  {(28)}^{3}

➖➖➖➖➖➖➖➖➖➖

HOPE YOU GOT IT

➖➖➖➖➖➖➖➖➖➖


Anonymous: .. loved to help you.. thanks
yash2090: welcome
pawan708721: but in competitive exam this takes lot of time
Answered by nilesh102
3

hi mate,

Answer:

(¡)5832

=>2×2×2×3×3×3×3×3×3

=> 18 ×18×18

=>(18)³

(¡¡) 1728

=>2×2×2×2×2×2×3×3×3

=> 12 ×12×12

=>(12)³

(¡¡¡) 216000

=>2×2×2×2×2×2×3×3×3×5×5×5

=> 60×60×60

=>(60)³

(iv) 21952

=>2×2×2×2×2×2×7×7×7

=> 28×28×28

=>(28)³

i hope it helps you.

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