Totally unrelated to Taxicab numbers

$n^3 < n^3+2n^2+9n+8 < (n+2)^3.$

Which implies that $n^3+2n^2+9n+8 = (n+1)^3.$

Thus, $n^2 = 6n+7,$ so $n=7.$

Firstly, Consider the equation $n^3 + 2n^2 + 9n + 8 = (n+2)^3 -4n^2 + 3n = a^3$ where $n$ is a positive integer. As all the terms are positive, this implies that $a^3$ is positive as is $a$ . If we rearrange the equation we can get $(n+2)^3 -4n^2 - 3n = a^3 \\ -3n - 4n^2 = a^3 - (n+2)^3 \\ -n(3+4n) = (a - n - 2)(a^2 + a(n+2) + (n+2)^2 )$ Notice that for positive $n$ , $3 +4n$ is always positive so the LHS is negative. If you also look at the RHS, $a^2 + a(n+2) + (n+2)^2$ is always positive for positive $a, n$ so therefore $a - n - 2$ is negative. That is $a - n - 2 < 0 \implies a < n-2$

Next, We can also state that $n < a$ because $n^3 < n^3 + 2n^2 + 9n + 8 = a^3$ . If we combine the two inequalities, we get the penultimate $n < a < n+2$ Since $a, n$ are integers, it only makes sense that $a$ is an integer between integers $n$ and $n+2$ so that can only leave $a = n+ 1$ . We can now substitute this result back into the original equation and the result follows. $n^3 + 2n^2 + 9n + 8 = (n+1)^3 \\ n^3 + 2n^2 + 9n + 8 = n^3 + 3n^2 + 3n + 1 \\ n^2 - 6n - 7 = 0 \\ (n-7)(n+1) = 0 \\ n = 7, n = -1$ Given that $n$ is positive, we can ignore our value of $n = -1$ so we are left with $\boxed{n = 7}$

First of all, let's consider:

$n^3 + 2n^2 + 9n + 8 = m^3$

Where $m$ is a positive integer.

As $n$ is a positive integer $(n>0)$ , it is clear that $n^3 + 2n^2 + 9n + 8 > n^3$ . Let's, then, compare the expression with $(n+2)^3$ :

$(n+2)^3 = n^3 + 6n^2 + 12n + 8$

Once that $n>0$ , $6n^2 + 12n > 2n^2 + 9n$ , and therefore:

$(n+2)^3 > m^3 > n^3$

$(n+2) > m > n$

From the inequation above, we can conclude that $m = n+1$ , so:

$(n+1)^3 = n^3 + 2n^2 + 9n + 8$

$n^3 + 3n^2 + 3n + 1 = n^3 + 2n^2 + 9n + 8$

$n^2 -6n -7 = 0$

The equation above has two trivial solutions: $n = 7$ and $n= -1$ . Yet, the problem asked for a positive integer, which means the final answer is $n = 7$ .

Hope it helped!

n^3+2n^2+9n+8=(n+1)(n^2+n+8)=a.a^2. n+1=a&n^2+n+8=a^2. n=a-1 by substitut. (a-1)^2+a-1+8=a^2 . a=8 &n=7######

We can compute this quite easily. Consider the following python code:

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cubes = [n**3 for n in range(100)]
for n in range(100):
    if n**3 + 2 * n**2 + 9 * n + 8 in cubes:
        print n

This results in 0 or 7, but obviously 0 is not a valid answer. So the answer is $\boxed 7$ .

In the equation, 8 is already a perfect cube: $2^3$

The next and smallest perfect cube is $(2^3)^3=2^9$ or $8^3$

If this is true, then $n$ is either a decimal or an integer.

Set the equation as $n^3+2n^2+9n+8=8^3$

Solve the equation, I did it by graphing.

$n$ turns out to be $7$