The other tutor cheated and used a computer, not algebra.

Let the hundreds digit be h, the tens digit be t, and the units digit be u
 
h + t² + u³ = 100h + 10t + u

t² - 10t +  u³ - u = 99h 

The number will be large as possible if the hundreds digit h is as large
as possible.

h will be as large as possible if the right side 99h is as large as possible.

The right side will be large as possible if the left side is as large as
possible.

The left side will be large as possible if the cube u³ is as large as
possible.  So we begin by trying u=9 and seeing if that can work.  If
not we'll try u=8, and so on down.  We consider it a quadratic in t
and use the quadratic formula:

         t² - 10t +  u³ - u = 99h
   t² - 10t +  u³ - u - 99h = 0
      t² - 10t + (u³-u-99h) = 0















If the + holds the square root must be less than or equal 4
If the - holds the square root must be less than 5

So 25-u³+99h+u must be less than or equal 25

We see if u=9 will work to make 25-u³+99h+u a positive perfect square
less than 25

25-u³+99h+u = 25-9³+99h+9 = 25-729+99h+9 = -695+99h

We set that between 0 and 25, inclusive.

0 <= -695+99h <= 25
695 <= 99h < 720
7.02... <= h <= 7.27...
That inequality does not allow h to be a digit, so u cannot be 9.

So we see if u=8 will work to make 25-u³+99h+u a positive perfect square
less than 25

25-u³+99h+u = 25-8³+99h+8 = 25-512+99h+8 = -479+99h

We set that between 0 and 25 inclusive

0 <= -479+99h <= 25
479 <= 99h < 504
4.83... <= h <= 5.09...
That says h must be 5

We see if that makes -479+99h a perfect square:

-479+99h = -479+99(5) = -479+495 = 16, yes for 16 is 42.

So we have u = 8,  h = 5, 

and we find t:  







We can use either the + or the -, but to make t as
large as possible we choose the +,

so t = 9

Answer:  598

Checking: 5 + 92 + 83 = 5 + 81 + 512 = 598

Edwin