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anonymous

  • one year ago

What is the difference in the total number of factors and the number of prime factors in 60^3 - 32^3 - 28^3

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  1. mathmath333
    • one year ago
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    \(60^{3}- 32^{3} - 28^{3} =3\times 32\times 28\times 60\)

  2. anonymous
    • one year ago
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    options : 1) 116 2) 120 3) 126 4) 130

  3. ganeshie8
    • one year ago
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    Start with : \(60-32=28\) cube both sides \((60-32)^3 = 28^3\) For left hand side, recall the identity \((a-b)^3 = a^3-b^3-3ab(a-b)\) : \(60^3-32^3 -3\cdot 60\cdot 32(60-32) = 28^3\) \(\implies 60^3-32^3-28^3 = 3\cdot 60\cdot 32\cdot 28=2^9\cdot 3^2\cdot 5\cdot 7\)

  4. ganeshie8
    • one year ago
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    so thats the prime factorization do you know how to find the number of factors of a number from prime factorization ?

  5. anonymous
    • one year ago
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    Yes.

  6. ganeshie8
    • one year ago
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    good, see if you can finish the rest

  7. anonymous
    • one year ago
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    Thank you :)

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