I do not quite understand what is meant by this question.

The term in the cube root reduces to the cube root of (a^99999 * b^66666), or in other words, a^33333 * b^22222.

This can be further rewritten down into (a^3 * b^2)^11111.

From the looks of things, a and b can be any two random numbers amd tje value of (a^3 * b^2 can be any possible value. But we are not certain of the divisibility of (a^3 * b^2) because this is highly dependent on our choice of a and b. Natural numbers are basically non-negative integers comprising 0 and all the whole numbers.

The only factor which we are certain of is 1 (if a and b are unknown), so the only way we can be certain that the product of the cube root and 1/n2 is when 1/n2 is not a fraction, or the largest integer value of n is 1.

Thanks for the reply. I was able to get to the same steps in terms of simplifying the term in the cube root. I then tried the following (not sure if it's right):

(a³³³³³ × b²²²²²) × (1 / n²)
= (a¹¹¹¹¹ × a²²²²² × b²²²²²) × (1 / n²)
= a¹¹¹¹¹ × [(ab)¹¹¹¹¹]² × (1 / n²)

So... since b > a, we can say that n² could be equal to [(ab)¹¹¹¹¹]², and n = (ab)¹¹¹¹¹

That's where I'm stuck. -_-"

Well spotted! I failed to realise that a and b are defined to be integers based on the question.

Then yes, you are correct.

Or even better: a^11111 can be broken further down into (a^5555)^2 times one more a.

So in short,

a^33333 * b^22222
= a^1 * a^33332 * b^22222
= a * a^16666 * a^16666 * b^11111 * b^11111
= a * (a^16666 * b^11111)^2

So n will be equal to the term inside this bracket (a^16666 * b^11111).

The idea will be better explained in Indices once you reach Secondary 3 (if you are in Express Stream or Normal Academic Stream).