A sum of cubes

We have $c = - (a+b)$ .

Therefore,

$a^3 + b^3 + c^3$

$= a^3 + b^3 - (a+b)^3$

$= a^3 + b^3 - (a^3 + b^3 + 3a^2b + 3ab^2)$

$= -3ab(a+b) < 1000$

So we have $-ab(a+b) < 334$ .

First, we note that $-ab(a+b)$ must be even. If $a$ or $b$ are even, then the term is even. If both are odd, then the sum $a+b$ (and therefore term) is even.

Now we look at $-ab(a+b) = 332 = 2 \times 2 \times 83$ . Since $a$ , $b$ and $c$ are integers, the terms $|a|$ , $|b|$ and $|a+b|$ must take values of $(1,1,332)$ , $(1,2,166)$ , $(1,4,83)$ or $(2,2,83)$ in any order, which isn't possible.

Now consider $-ab(a+b) = 330$ . $(a,b)=(-5,-6)$ works, therefore we have our answer as $3 \times 330 = 990$ .

Let the largest integer be $N$ .

We have $N=a^3+b^3+c^3$ where $a+b+c=0$

Now, let's try to factorize the left hand side:

1) Add and deduct $3abc$

$\Rightarrow N= a^3+b^3+c^3-3abc+3abc$

2) Now we use the fact that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Thus, our equation reduces to: $N=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$

Now, since it is given that $a+b+c=0$ , we have:

$N=3abc$

Now, $N\leq 999$

$3abc\leq 999$

$\Rightarrow abc \leq 333$ .

Now, we first check if there exists positive integer values of $N$ , such that the conditions given are satisfied. If there exists such positive $N$ , then obviously we wouldn't consider negative $N$ or $N=0$ .

Since we first take $N$ as a positive integer, we must have that $a,b,c$ are all positive or either two are negative and one positive. The former case will never yield $a+b+c=0$ , so we need to have one positive integer and the other two negative. Taking for instance $b=-x$ and $c=-y$ for some positive integers $x$ and $y$ , our condition becomes $a-x-y=0$ . As such, we need to find triples $(a,x,y)$ , so our previous equation is satisfied. We must keep in mind that since we check for positive $N$ , we have that $(a,b,c \neq 0)$ . [Note: Since, now we have $a=x+y$ , we have $a>\mid b \mid ,a>\mid c \mid$ ]

Now in order to maximize $N$ , we maximize $abc$ .

We will do some case checking starting with largest possible value of $abc$ , since we need to maximise the product $abc$ .

1)When $abc=333$ :

The prime factorization of $333=3^2 \cdot 37$

The possible triples are: $(37,9,1),(37,3,3)$

None of these satisfy $a-x-y=0$ .

2) When , $abc=332$ :

Note that $332=2^2 \cdot 83$

As such we can have the following possible triples: $(83,4,1),(83,2,2),(146,2,1)$

Clearly, none of these satisfy $a-x-y=0$

2) When $abc=331$ :

Now since $331$ is prime and the only possible triple $(331,1,1)$ will not satisfy our condition. So, we ignore this case as well.

3)When $abc=330$ :

We can write, $330=2 \cdot 5 \cdot 11 \cdot 3$ .

We find that the pair $(11, 5, 2\cdot 3)=(11,5,6)$ satisfies $a-x-y=0$ where $a=11,b=-5,c=-6$ .

Thus, the maximum value of $abc=330$ for $abc \leq 333$ , $a+b+c=0$

For which we have $N=990$ and we are done.

Given that a+b+c is = o .So a^3 + b^3 + c^3 = 3 a b c . So , 3 a b c < = 1000 or a b c <= 333 .Since a , b, c are integers . Now , 333 = 3 * 3 * 37

332 = 83 * 2 * 2 ,

331 = 331 * 1 ,

330 = 6 * 5 * 11 ,

or 330 = - 6 * - 5 * 11

As 11 - 6 - 5 = 0 So , a = - 6 b = - 5 c = 11 Hence 3 a b*c = - 6 * - 5 * 11 * 3 = 990 Hence the largest integer less than 1000 that can be expressed as a^3 + b^3 + c^3 is 990 .

Since a + b + c =0. We got

\sq( a + b + c )^2=0 or

a ^3+ b ^3+ c ^3+3 a b ^2+3 b c ^2+3 c a ^2+3 a ^2 b +3 b ^2 c +3 c ^2 a +6 abc =0

But since a + b + c =0, a + b =- c and thus,

a ^3+ b ^3+ c ^3+6 abc -3 abc -3 abc -3 abc =0.

this is equivilant to

a ^3+ b ^3+ c ^3-3 abc =0

But by AM-GM inequality we get a = b = c which is impossible.

So, we deduct that at least the sum of 2 number is equal to the another number. Without loss of generality we assume a + b = c . Thus, by using inequality we get a ^3+ b ^3< c ^3

Thus, when (-5, -6, 11), the equation achieves its maximum value, which is 990.

$a+b+c=0$

$a+b=-c$

$(a+b)^3 = (-c)^3$

$a^3+ 3a^2 b+ 3a b^2 +b^3 = -c^3$

$a^3+ 3ab(a+b) +b^3=-c^3$

Subtitute $a+b=-c$

$a^3+3ab(-c)+b^3+c^3=0$

$a^3+b^3+c^3=3abc$

So, when $a+b+c=0$ then $a^3+b^3+c^3=3abc$

Since we must find the largest integer that can be represented as $3abc$ , it must be a multiple of $3$

The list of possible integer in descending order is $999,996,993,991,987,...$

$999=3^3 \times 37, 996=2^2 \times 3 \times 83, 993= 3 \times 331, 990= 2 \times 3^2 \times 5 \times 11, 987= 3 \times 7 \times 47,...$

To make $3abc$ positive, all the factor must be positive or two of the factors must be negative. Since $a+b+c=0$ , it is not possible for all the factors to be positive therefore two of the factor must be negative

So, the largest integer possible is $990$ when $a=11, b=-6, c=-5$

Because a+b+c=0, so c=-a-b.
We have: a^3+b^3+c^3 = a^3+b^3+(-a-b)^3 = -3ab^2 -3ba^2 = -3ab(a+b) = 3abc.
Because a^3+b^3+c^3 < 1000, so abc<334. We use prime factorization: 333=3^2 * 37, because 37>3^2, so 333 cannot be the value of abc.
332=2^2 * 83, because 83>2^2, so 332 cannot be the value of abc. 331 is a prime number and cannot be the value of abc. 330=2 * 3 * 5 * 11, and 2*3 + 5 = 11, so abc = 330 and the number that satisfy the problem is 990 = 11^3 + (-5)^3 + (-6)^2

since we a+b+c has to be zero therefore a^3+b^3+c^3 must be equal to 3abc. so for abc to be positive, two out of a,b,c must be negative. there fore the value of 3abc would be just less than 1000 when a=11,b=-5,c=-6.therefore , a^3+b^3+c^3=3 into 11 into -5 into -6=990.

Since $a+b+c = 0$ , we also have $c = -(a+b)$ . Substituting into the original equation, we get $a^3 +b^3+c^3 = -3ab(a+b)$ . To get this expression positive, either $a$ or $b$ must be negative. We can assume without loss of generality that $b$ is negative. In this case, set $b = -d$ to get a simpler problem involving only positive integers:

$a^3 + b^3 + c^3 = 3ad(a-d)$

To get this expression close to but less than $1000$ , we have to get $ad(a-d)$ close to but less than $333$ . A quick search reveals that it is not possible to have the numbers $333, 332, 331$ in this form. But since $330 = 11 \cdot 6 \cdot 5 = 11 \cdot 6 \cdot (11-6)$ , we can set $a = 11, d = 6$ such that $3ad(a-d) = 990$ , our final answer.

If we want a large number, we want one of a,b,c to be a large positive number and the other two to be smaller negatives so that a+b+c=0. (If we were to do two small positives and one large negative, a^3+b^3+c^3 would be a negative number. We also cannot set one variable equal to zero, or else a^3+b^3+c^3 = 0)

I chose to make the variable "a" as my "large positive number". a = -(b+c) a^3 = -(b^3 + 3b^2c^1 + 3b^1c^2 + c^3) a^3 + b^3 + c^3 = -3b^2c^1 + -3b^1c^2 = -3bc(b+c) [note that this will be a positive value if b and c are negative]

All right, we want the largest value of -3bc(b+c) less than 1000 such that both b and c are integers. [Note: I already factored out a 3 below]

First, we check if 333 = -bc(b+c) for integers b and c 333 = 9 37 (which cannot be made with -bc(b+c)) Next, we check if 332 = -bc(b+c) 332 = 4 83 (still not solvable) 331 is prime, so no luck there. 330 = 2 3 5 11 = 11 30 (which works if b= -5 and c= -6) Thus, our solution is -3bc(b+c) = 990.

From $a+b+c=0$ , we see that either one or two of the integers $a,b,c$ are positive and the other/s is/are negative, or all three are $0$ . Because $a^3+b^3+c^3$ can certainly be made greater than $0^3+0^3+0^3=0$ (just consider $a=3,b=-2,c=-1$ ), we discard the $a=b=c=0$ case.

What happens if two are positive and the other one is negative? Without loss of generality, let $a,b>0$ and $c=-(a+b)<0$ . $a^3+b^3+c^3 =$ $a^3+b^3+(-(a+b))^3=$ $a^3+b^3-a^3-3a^2b-3ab^2-b^3=$ $-3(a^2b+ab^2)<0$ since $a,b>0$ . Again, because we can clearly make $a^3+b^3+c^3$ positive, we discard this case.

We're left with the case where, without loss of generality, $b,c<0$ and $a=-(b+c)>0$ . Through algebra similar to that seen in the previous case, we have $a^3+b^3+c^3=$ $-3(b^2c+bc^2)=$ $-3bc(b+c)>0$ .

Thus, we're trying to find the largest integer less than $1000$ which can be expressed as $3bc(b+c)$ with integers $b,c>0$ (this is equivalent to $-3bc(b+c)$ with integers $b,c<0$ ). Clearly, we only need to check multiples of $3$ :

$999=3^3\cdot37$

$996=2^2\cdot3\cdot83$

$993=3\cdot331$

None of these can be expressed in the desired form. However, $990=2\cdot3^2\cdot5\cdot11=$ $3\cdot5\cdot6\cdot11$ can be made using $b=5,c=6$ . Indeed, $a=11,b=-5,c=-6$ gives $a^3+b^3+c^3=$ $11^3+(-5)^3+(-6)^3=$ $1331-125-216=990$ .

Therefore, the largest integer less than $1000$ which can be expressed as $a^3+b^3+c^3$ with $a,b,c\in\mathbb{Z}$ is $\fbox{990}$ .

From given, $c=-(a+b)$ ; plugging in, we want to maximize the quantity $a^3+b^3+(-a-b)^3$

$a^3+b^3-a^3-b^3-3ab(a+b)$

$=-3ab(a+b)$

Of course the maximum of this quantity is positive; we can write $a=-c$ and $b=-d$ , and now we are maximizing $3cd(c+d)$ . This quantity is, by the given restriction, less than 1000, so $cd(c+d)$ must be less than or equal to 333, and to find the original value we take this value and multiply by 3. Checking each case, working downwards, we see that this can not attain a value of 333, 332 or 331, but 330 can be obtained with c=5 and d=6 (and permutations), so the answer is $3*5*6*11=990$ .

The solution will be easy to determine if you find the number first and then find the match sign (positive or negative).

Because $a+b+c=0$ , then one of the variable should be the difference from each other. Assume that $a$ as the largest one and $c=a-b$ , it will form $a-b-(a-b)=0$ , then raise variable power to $3$ , it must $a^{3}-b^{3}-(a-b)^{3}$ .

Expand $a^{3}-b^{3}-(a-b)^{3}$

$3a^{2}b-3ab^{2}=3ab\times(a-b)$

Then it form $3ab\times(a-b)<1000$ , now we know that the integer must be divisible by $3$ .

Because the problem wants the largest integer as solution, therefore I take the sample from large number.

There was $999$ , $996$ , $993$ and $990$ (and more but I don't list them because they out of solution), then I try to match them one by one.

Fortunately, $(990,-990)$ can be form with $a=(11,-11)$ , $b=(-6,6)$ , $c=(-5,5)$ .

So, $a^{3}+b^{3}+c^{3}>1000\Rightarrow11^{3}+(-6^{3})+(-5^{3})=990$

Because $a+b+c=0,$ one can prove that $a^3+b^3+c^3=3abc$ .

So if $n=a^3+b^3+c^3$ , it is divisible by $3$ . Also, two of the numbers $a,b,c$ must be negative, and one positive. Without loss of generality, we can assume that $a=-m,\ b=-k,\ c=m+k$ where $m$ and $k$ are positive integers. So $n=3mk(m+k)$ . Note that $mk(m+k)$ is always even, and it must be at most $333$ . If $mk(m+k)=332=2^2\cdot 83,$ then one of the numbers $m,k,m+k$ must be at least $83$ , which is too big for the other numbers. So the largest $mk(m+k)$ can possibly be is $330$ . Note that $330=5\cdot 6\cdot 11$ , which gives $990=(-5)^3+(-6)^3+11^3.$ Therefore, the answer is $990.$

We can rearrange $a+b+c=0$ to become $a=-(b+c)$ .

Substituting this into $a^3+b^3+c^3$ , we get $-(b+c)^3+b^3+c^3$ , which simplifies to $-3bc^2-3b^2c$ .

We can look at this as a quadratic in $c$ , with $b$ initially considered at a constant. Since the coefficient for $c^2$ is negative, the expression has a maximum at $c=\frac{-(-3b^2)}{2(-3b)}$ , or $c=\frac{-b}{2}$ .

Plugging this value into $-3bc^2-3b^2c$ , we get $\frac{-3b^3}{4}+\frac{3b^3}{2}$ , or $\frac{3b^3}{4}$ .

Since the expression must be less than 1000, we get that $\frac{3b^3}{4} < 1000$ , or that $b^3<\frac{4000}{3}$ . Since $\frac{4000}{3} \approx 1333.33$ , the maximum possible value for $b$ is $11$ , since $11^3=1331$ .

Plugging this back into $-3bc^2-3b^2c$ , we get $-33c^2-363c$ . This expression is maximized at $c=\frac{-363}{2(-33)}$ , which is $c=\frac{-11}{2}$ .

However, $c$ must be an integer, so we try the integers closest to $\frac{-11}{2}$ , $-5$ and $-6$ . Plugging in each of these two values for c yields $990$ , so that is our answer.

$a+b+c=0$ $(a+b+c)^3=0$

This problem gives us two equations:

$\begin{cases} a^3 + b^3 + c^3 < 1000 \\ a + b + c = 0 \end{cases}$

Rearranging the second equation we get $c = -a - b$

If we substitute that into the first equation, it simplifies the equation down to $-3a^2b - 3ab^2$

Assuming $b$ is positive and known, we can reformat it to $-(3b)a^2 + (-3b^2)a < 1000$

Since this is a downwards-facing parabola, we can say it has a maximum at $\frac {-(-3b^2)} {(-6b)} = \frac {-b} {2}$

If we substitute this in for $a$ , we get

$\frac{3b^3} {4} < 1000$

$3b^3 < 4000$

$b^3 < 1333.\overline{3}$

The closest cube less than $1333.\overline{3}$ is $1331$ which will make $b$ equal to $11$ .

Since we now know $b$ we can solve for $a$ . If we plug $b$ back in, we will get the quadratic $-33a^2 - 363a < 1000$ . If we solve for the vertex, we find that $a = \frac{363} {-66} = -5.5$ .

$-5.5$ is between two integers, $-5$ and $-6$ . However, quadratic equations are symmetrical around the vertex, so both integers should yield the same result. In fact,

$\begin{cases} -3(-5)^2(11) - 3(-5)(11)^2 = -825 + 1815 = \fbox{990} \\ -3(-6)^2(11) - 3(-6)(11)^2 = -1188 + 2178 = \fbox{990} \end{cases}$

$990$ is the answer.