## 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

1. mathmath333

$$60^{3}- 32^{3} - 28^{3} =3\times 32\times 28\times 60$$

2. anonymous

options : 1) 116 2) 120 3) 126 4) 130

3. ganeshie8

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

4. ganeshie8

so thats the prime factorization do you know how to find the number of factors of a number from prime factorization ?

5. anonymous

Yes.

6. ganeshie8

good, see if you can finish the rest

7. anonymous

Thank you :)