Let a and b be relatively prime integers with a b 0 and

Question

Let a and b be relatively prime integers with a > b > 0 and  \frac{a^3 −b^3}{(a−b)^3}=\frac{73}{3}  What is a − b?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

Krunaal · Accepted Answer

You actually don't need to simplify this. We are told that  a , and  b  are prime integers. So,  a - b  will be an integer; similarly,  a^3 - b^3  will be an integer.

Now,  \frac{a^3 −b^3}{(a−b)^3}=\frac{73}{3}

=>  a^3 −b^3=\frac{73}{3} * (a-b)^3

For  a^3 - b^3  to be an integer  (a-b)^3  needs to be a multiple of 3, so that the 3 in the denominator can be cancelled out and we get an integer.

If you glance through the answer choices, only option C. fits the criteria.

Abhishekgmat87 · Answer

There is 3 in denominator in RHS and cube in LHS. Convert 3 to 27. Multiply numerator by 9, it becomes 657 which is equal to 10^3- 7^3

a=10, b= 7

Option C

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sanjitscorps18 · Answer

a^3−b^3/(a−b)^3=73/3

(a-b)(a^2 + ab + b^2) / (a−b)^3 = 73/3

=> (a^2 + ab + b^2) / (a−b)^2 = 73/3

=> 1 + 3ab/(a−b)^2 = 73/3

=> 3ab/(a−b)^2 = 70/3

=> 9ab = 70(a−b)^2

Now 70 has to be a factor of ab since a and b are co-primes and cannot be a square value. Hence (a−b)^2 = 9
=> a-b = 3

Option C

KChop · Answer

Could someone please explain how this is simplified

Let a and b be relatively prime integers with a > b > 0 and

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