If x and y are positive integers such that x³– y³ = xy + 61, then find the value of (x + y).
Answers
Given : x and y are positive integers such that x³– y³ = xy + 61,
To find : value of (x + y).
Solution:
x³– y³ = xy + 61
x > y
x & y both can not be even as
then LHS = even & RHS = odd
x ≥ 4 as x³– y³ > 61
Case 1
x : 4 then xy > 4 Hence RHS > 65
so not possible
Case 2
x = 5
y = 1
=> LHS = 124 , RHS = 66
y = 2 =>
LHS = LHS = 117 , RHS = 71
y = 3
=> LHS = 98 , RHS = 76
y = 4
=> LHS = 61 , RHS =81
Case 3 :
x = 6
y = 1
=> LHS = 215 RHS = 67
y = 3
LHS = 189 RHS = 79
y = 5
LHS = 91 RHS = 91
x = 6 & y = 5 Satisfy this
x + y = 6 + 5 = 11
as this solution has difference between x & y as only 1
so no further solution possible
as x³– y³ > xy + 61 for all possible values of x > y
Lets check x = y + 1
=> ( y + 1)³ - y³ = (y + 1)y + 61
=> y³ + 1 + 3y² + 3y - y³= y² + y + 61
=> 2y² + 2y - 60 = 0
=> y² + y - 30 = 0
=> (y + 6)(y - 5) = 0
=> y = -6 or y = 5
Hence another possible solution
x = - 5 , y = - 6 ( but negative integers)
so only possible values of x & y
are 6 & 5
x + y = 11
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