Question 1208880
<font color=black size=3>
The other tutor has a great solution. 
Perhaps it's the most efficient pathway to solve. 
I'll provide another method.


w = a+b
(a+1)(b+1)(a+b) = 1530
(ab+a+b+1)(a+b) = 1530
(ab+w+1)w = 1530
ab+w+1 = 1530/w
ab = 1530/w - w - 1
Let's call this equation (1) to use later. 


Then notice how,
w = a+b
w^2 = (a+b)^2
w^2 = a^2+2ab+b^2
w^2-3ab = a^2+2ab+b^2-3ab
w^2-3ab = a^2-ab+b^2
Let's call this equation (2)


Use the sum of cubes formula to say the following
a^3+b^3 = (a+b)(a^2-ab+b^2)
a^3+b^3 = w(w^2-3ab) ............................. plug in equation (2)
1241 = w(w^2-3*( 1530/w - w - 1 )) ............... plug in equation (1)
1241 = w^3-3w*( 1530/w) + 3w^2 + 3w
1241 = w^3-4590 + 3w^2 + 3w
w^3-4590 + 3w^2 + 3w-1241 = 0
w^3 + 3w^2 + 3w - 5831 = 0
From here you can use a graphing calculator to graph y = x^3 + 3x^2 + 3x - 5831
The only real number root is x = 17. This is where the cubic curve crosses the x axis. You'll likely need to adjust your window to be able to see this root.
As a check,
x^3 + 3x^2 + 3x - 5831 = 0
(17)^3 + 3(17)^2 + 3(17) - 5831 = 0
4913 + 867 + 51 - 5831 = 0
0 = 0
This proves that w = 17 is a solution to w^3 + 3w^2 + 3w - 5831 = 0


Through trial and error (aided with a spreadsheet), you can determine that a = 8 and b = 9 satisfy the two given equations.<table border = "1" cellpadding = "5"><tr><td>(a+1)*(b+1)*(a+b) = 1530</br>(8+1)*(9+1)*(8+9) = 1530</br>9*10*17 = 1530</br>1530 = 1530 .... works</br></td><td>a^3+b^3 = 1241</br>8^3+9^3 = 1241</br>512+729 = 1241</br>1241 = 1241 .... works</br></td></tr></table>So we conclude that a+b = 8+9 = 17
Due to symmetry, we could also say that a = 9 and b = 8.



Answer: <font color=red>17</font>
</font>