Answer:
3,4,5
Step-by-step explanation:
Here, we need to asume that the values are integers, as if not, there will be infinite solutions.
Lets go by parts:
The 1st number is a cubic, lets call it A
The second is a factor of 20, lets call it B. The same for the 3rd, call it C.
So, A + B+ C = 36
There are no many factors of 20, those are the numbers that when multiplied gives as the value of 20. Those are 1, 2, 4, 5, 10 and 20. So, B and C are some of these numbers.
Then, we know A if less than 36, other way the whole sum will be greater than 36. How many cubic number with integer cubic roots are less than 36? Well, lets guess:
1^3 = 1 -> valid, as it is less than 36
2^3 = 8 -> valid, as it is less than 36
3^2 = 27 -> valid, less than 36
4^3 = 64 -> not valid, as it is greater than 36
So, A ir 1, 2 or 3.
If A is 1, then B+C needs to be 35. But, from the factors of 20 that we listed, there are no combinations of 2 numbers that sum 35. So, A CAN'T be 1.
If A = 2, then we have:
8 + B + C = 36
B + C = 28
And again, there are not combinations of two factors of 20 that sum 28 (try yourself).
If A=3:
27 + B + C = 36
B +C = 9
And we have a winner!!! If B=4 and C=5 (or viceversa C=4 and B=5)
27 + 4 + 5 = 36
So, A=3, B=4 and C=5 or A=3, C=4 and B=5 are solutions.
