Answer:

Step-by-step explanation:

The first thing anyone would try is  x=3,y=4 , which checks. It will be useful to have one root so we can turn our cubic equation into a QUADRATIC one.

Proceeding like Mr. Pradeep, let  s=x+y . I USE  s  because my  t s look like  +  signs. Of course my  s  looks like a  5  ... anyway.

s2=(x+y)2=x2+2XY+y2=25+2xy  

2xy=s2−25  

s3=(x+y)3=x3+3x2y+3y2x+y3=91+3xy(x+y)=91+3sxy  

0=−2s3+3s(2xy)+182=  −2s3+3s(s2−25)+182=s3−75s+182  

Since  3,4  is a solution,  s=3+4=7  must be a root of this equation, so we can factor out  (s−7).  Doing the long division, we get

s3−75s+182=(s−7)(s2+7s−26)=0  

The THREE roots are  s=7  and  s=−7±317√2 . The latter two are real and real ugly. I can see why Mr. Pradeep resorted to decimal approximations.

Now that we have  s , we can go back and solve for  x  and  y . Let's do it symbolically first.

We have  s=x+y , so  y=s−x , and  2xy=s2−25 .

2x(s−x)=s2−25  

0=2x2−2sx+(s2−25)  

x=14(2S±(2s)2−4(−2)(s2−25)−−−−−−−−−−−−−−−−−−√)=12(s±50−s2−−−−−−√)  

So for  s=7  we have  x=12(7±50−72−−−−−−√)=12(7±1) .  x=3  or  x=4 . Note that when one is  x  the other is  y .

Now we're left with the messier parts. I'm busy now so I will come back to those.

10 days later and I'm back. This is messy, but let's get those other roots. We have

s=−7±317√2  and  x=12(s±50−s2−−−−−−√)  

That's enough to calculate the answers, but I was after the closed form. It doesn't simplify down much, but we can make progress.

(2s)2=49+9⋅17−±2⋅2117−−√=202−±4217−−√  

200−4s2=−2±4217−−√  

We're going to have to take the square root of that. Let's first focus on the  s  with the plus sign, which gives a positive  200−4s2 .

250−s2−−−−−−√=200−4s2−−−−−−−−√=−2+4217−−√−−−−−−−−−−√  

x=12(s±50−s2−−−−−−√)  

x=  14(−7+317−−√±−2+4217−−√−−−−−−−−−−√)  

Simple. These give two more real solutions,  x≈4.613  and  x≈−1.928.  Again, each is the other's  y .

What about that complex solution? That's from  s=−7−317√2  

x=  14(−7−317−−√±i2+4217−−√−−−−−−−−−√)  

Those are two more complex roots,  x≈−4.842±3.309i  

OK, we found six roots. They come as three  x,y  pairs that you can swap due to the symmetry of the problem.