Noshad wrote:
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
a and b are relatively prime and a>b>0.
\(\frac{a^3 −b^3}{(a−b)^3}\) =\(\frac{73}{3}\)
\(\frac{(a-b)(a^2 +ab + b^2)}{(a−b)^3}\) =\(\frac{73}{3}\)
Since a>b>0 So, a-b=/=0
\(\frac{(a^2 + ab + b^2)}{(a−b)^2}\) =\(\frac{73}{3}\)
\(\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}\) =\(\frac{73}{3}\)
\(\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}-1\) =\(\frac{73}{3}-1\)
\(\frac{(3ab)}{(a^2 - 2ab + b^2)}\) =\(\frac{70}{3}\)
\(\frac{(ab)}{(a-b)^2}\) =\(\frac{70}{9}\)
\(\frac{(ab)}{(a-b)^2}\) =\(\frac{70x^2}{(3x)^2}\)
so, a-b = 3x and ab = 70x^2
At x =1, a-b = 3, ab = 70 -> (3+b)b= 70 ->b^2 +3b -70 = 0 -> (b+10)(b-7) = 0 -> b= 7, a = 10 , a-b = 3
At x= 2, a-b = 6 , ab = 70*2^2, a= 20, b=14
and so on... But a and b are co-prime and hence a-b =3 , ab = 70, a = 10 & b=7.
Answer CThanks a lot..