anonymous
  • anonymous
What is the difference in the total number of factors and the number of prime factors in 60^3 - 32^3 - 28^3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(60^{3}- 32^{3} - 28^{3} =3\times 32\times 28\times 60\)
anonymous
  • anonymous
options : 1) 116 2) 120 3) 126 4) 130
ganeshie8
  • ganeshie8
Start with : \(60-32=28\) cube both sides \((60-32)^3 = 28^3\) For left hand side, recall the identity \((a-b)^3 = a^3-b^3-3ab(a-b)\) : \(60^3-32^3 -3\cdot 60\cdot 32(60-32) = 28^3\) \(\implies 60^3-32^3-28^3 = 3\cdot 60\cdot 32\cdot 28=2^9\cdot 3^2\cdot 5\cdot 7\)

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ganeshie8
  • ganeshie8
so thats the prime factorization do you know how to find the number of factors of a number from prime factorization ?
anonymous
  • anonymous
Yes.
ganeshie8
  • ganeshie8
good, see if you can finish the rest
anonymous
  • anonymous
Thank you :)

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