What are two consecutive numbers whose cubes differ by 169

Let us assume the first number = x
Then
The consecutive number will be = x + 1
(x + 1)^3 - x^3 = 169
x^3 + 3x^2 + 3x + 1 - x^3 = 169
3x^2 + 3x = 169 - 1
3x^2 + 3x = 168
3x^2 + 3x - 168 = 0
Dividing both sides by 3 we get
x^2 + x - 56 = 0
x^2 -7x + 8x - 56 = 0
x(x - 7) + 8 (x - 7) = 0
(x - 7) (x + 8) = 0
From the above solution we have to take
x - 7 = 0
x = 7
So the small integer is 7
The larger integer = x + 1
                             = 7 + 1
                             = 8
So the two consecutive integers are 7 and 8.

If you divide 0.1 and 0.6 you wii get 0.166666 which rounds to .167 which is answer b