Asked by Meya22
                How do you find the sum of x³+y³+z³=k, for each k from 1 to 100.
Have a hard time figuring it out, and I have a 97-100 in Algebra such a shame.
Any one I would love to get help!!
            
        Have a hard time figuring it out, and I have a 97-100 in Algebra such a shame.
Any one I would love to get help!!
Answers
                    Answered by
            mathhelper
            
    I will assume that k will be a whole number,
Not all values of k from 1 to 100 can be obtained with whole numbers of
x, y , and z
I will also assume that we are staying away from negative values of x, y, z
If y^3 and z^3 are zero, then the largest x possible for x^3 ≤ 110 is x = 4
(4^3 = 64 and 5^3 = 125, x = 5 is too big)
The same is true for y if both x and z are zero etc
x^3 + y^3 + z^3 is symmetric, that is, for any given value of x, y , and z
we could interchange them to get the same result
e.g. 2^3 + 3^3 + 1^3 = 3^3 + 1^3 + 2^3 =
So all we need is 3 cubes whose sum ≤ 100
they are:
possible triples: each time checking if the sum of cubes ≤ 100
1 0 0 --- 3 of them
1 0 1 --- 3 of them
1 0 2 --- 3 of them
1 0 3 --- 3 of them
1 0 4 --- 3 of them
1 1 0 --- 3 of them
1 1 1 --- 1 of them
1 1 2 --- 3 of them
1 1 3 --- 3 of them
1 1 4 --- 3 of them
1 2 0 --- 3 of them
1 2 2 --- 3 of them
1 2 3 --- 6 of them
1 2 4 --- 6 of them
1 3 0 --- 6 of them
1 3 3 --- 3 of them
1 3 4 --- 6 of them
2 2 2 --- 1 of them
2 2 3 --- 3 of them
2 2 4 --- 3 of them
2 3 3 --- 3 of them
3 3 3 --- 1 of them , I count a total of 72
I avoided duplicates,
e.g. skipped 2 2 1 near the end since already had it at 1 1 2
I hope I got all of them, tried to be systematic about it.
check if I missed any.
If the question was to actually get the sum of x^3 + y^3 + z^3,
you would have a lot of tedious busy-work ahead, don't think that is
what was asked.
    
Not all values of k from 1 to 100 can be obtained with whole numbers of
x, y , and z
I will also assume that we are staying away from negative values of x, y, z
If y^3 and z^3 are zero, then the largest x possible for x^3 ≤ 110 is x = 4
(4^3 = 64 and 5^3 = 125, x = 5 is too big)
The same is true for y if both x and z are zero etc
x^3 + y^3 + z^3 is symmetric, that is, for any given value of x, y , and z
we could interchange them to get the same result
e.g. 2^3 + 3^3 + 1^3 = 3^3 + 1^3 + 2^3 =
So all we need is 3 cubes whose sum ≤ 100
they are:
possible triples: each time checking if the sum of cubes ≤ 100
1 0 0 --- 3 of them
1 0 1 --- 3 of them
1 0 2 --- 3 of them
1 0 3 --- 3 of them
1 0 4 --- 3 of them
1 1 0 --- 3 of them
1 1 1 --- 1 of them
1 1 2 --- 3 of them
1 1 3 --- 3 of them
1 1 4 --- 3 of them
1 2 0 --- 3 of them
1 2 2 --- 3 of them
1 2 3 --- 6 of them
1 2 4 --- 6 of them
1 3 0 --- 6 of them
1 3 3 --- 3 of them
1 3 4 --- 6 of them
2 2 2 --- 1 of them
2 2 3 --- 3 of them
2 2 4 --- 3 of them
2 3 3 --- 3 of them
3 3 3 --- 1 of them , I count a total of 72
I avoided duplicates,
e.g. skipped 2 2 1 near the end since already had it at 1 1 2
I hope I got all of them, tried to be systematic about it.
check if I missed any.
If the question was to actually get the sum of x^3 + y^3 + z^3,
you would have a lot of tedious busy-work ahead, don't think that is
what was asked.
                    Answered by
            Bosnian
            
    In google paste:
cuemath algebra/sum-of-cubes-of-n-natural-numbers/
When you see list of results click on;
Sum of Cubes of n Natural Numbers - Formula, Proof, Examples
You will find very detail explanation.
    
cuemath algebra/sum-of-cubes-of-n-natural-numbers/
When you see list of results click on;
Sum of Cubes of n Natural Numbers - Formula, Proof, Examples
You will find very detail explanation.
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