a and b are natural odd numbers. Find the last digit of

We are looking at units digit of \(a^3+b^3\)

1) Product of a and b ends with 1
a and b both can end with 1 as 1*1=1, and the units digit of \(a^3+b^3\) will be 1+1=2.
a and b can end with 9 and 9 as 9*9=81, and the units digit of \(a^3+b^3\) will be 9+9=18 or 8.
Also a and b can be 3 and 7.
Insuff

2) Sum of a and b is a multiple of 5
Now a and b are natural odd numbers, so a+b will be even, and 'a multiple of 5' means a+b is a multiple of 10.
Possible values..
The units digit of a+b=5+5=10....and the units digit of \(a^3+b^3=5^3+5^3\) will be 5+5=10 or 0.
The units digit of a+b=3+7=10....and the units digit of \(a^3+b^3=3^3+7^3\) will be 7+3=10 or 0.
The units digit of a+b=1+9=10....and the units digit of \(a^3+b^3=1^3+9^3\) will be 1+9=10 or 0.
In every case, the units digit of \(a^3+b^3\) is 0.
Sufficient

B
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a and b are natural odd numbers. Find the last digit of the sum of their cubes.

Stat 1: Product of a and b ends with 1
a and b can be (1,1), (9,9) or (7,3), Now, last digit of their cubes can't be same. Not sufficient

Stat 2:Sum of a and b is a multiple of 5
a and b can be (5,5) or (7,3), ow, last digit of their cubes will be same. Sufficient

So, I think B.