a⁴ + b⁴ + c⁴=?

 All right so,simultaneous equations... two main methods to deal with these: one is substitution and the other one is elimination. Now I did actually try this one out with substitution but it quickly got very messy; you don't want to use substitution for this one. Instead we're just going to sort of take each equation... 

I'm going to call this equation one, equation two and three, and we'll just see what we can get out of this. So the first thing that I thought of was, what if we squared both sides of equation one? Because then we're going to get some "a squared"s and "squared"s and maybe that's going to be able to be combined with equation two some how. So let's work out equation one squared.

 So this may not exactly be the same as elimination method that you're familiar with but it's basic ally the same sort of thing. Okay so square both sides of this. On the left, if you think of it like... if you're having trouble expanding it, then maybe write it out like this first. So we've got a squared from there and then  you'll have an ab, then there'll also be an ab from those and, you know, there's gonn abe bc and bc again and so on, so you're gonna get a squared + b squared +c squared and then there's two of each of the cross terms, I guess. Okay well that's really good because then we can subtract equation two from that and that's going to get rid of that,so I might just cross that out and change that to six on the right, so if wejust subtract that from both sides.

 And then let's divide both sides by 2, so that gives us ab + bc + ac is 3, so that might be useful. Now another thing we could do is we could multiply equation 1 by equation 2 because that's going to give us like an a cubed and b cubed and so on, so we might be able to combine that with equation 3 somehow. So equation 1 times equation 2:on the left we've got a + b + c times a²  + b²  + c² .That looked like a 4. And that equals that times that.what is equation

 So when we expand this we're going to get like the a cubed from there and the b cubed and so on and then we're also gonna get ab² plus ac² from there, and we're gonna get a ba²,whoops ba², plus bc² and a ca² and cb². So that's a bit annoying, but at least now we can subtract equation 3 from that. So if I just subtract equation 3,get rid of that, and subtract 22 from there, so we get that that is 18, so that might be useful later on. Now what else we could do is with this equation we might be able to do something with that. I might call that equation 4, and then we could try something like equation 1 times equation 4 and see what happens.With this one I just sort of tried a lot of things and not everything washelpful,

 but most of it was. One of the things I tried that wasn't helpful I think was I tried multiplying equation 1 by equation 3. There might be a way to do it with that but I didn't... it didn't lead to anything for me. Let's just try oute verything we can think of. So equation 1 times equation 4, so on the left we've got the a + b + c times that and then on the right we've got the 4 times 3, so 12.

 So that is going to make a²b plus...we'll have an abc plus a²c plus ab² plus b² another abc plus another abc from there, plus bc² andac² from there, and I'm running out of space again. Okay so we've got 3abc from those and then plus the rest of this stuff, which is actually the same as whatwe had here, right? what is equation Because we've got the a²b which was there and a²cwhich was there and so on, and we know that all of that is 18, 

sowe've got 3abc + 18 = 12, so that's good because we can work out abc.If we shift that over we get a negative 6 and then divide by 3 and we getneg ative 2, so that might come in handy. Another thing that we could do is we could square equation 4, see what we can come up with with that. So equation 4squared, so that squared. You're gonna get like the square of each one and thenplus the cross terms, and there's gonna be two of each, like because it's pretty much the same as squaring equation 1, if you remember that had the squares andthen 2 times the cross terms.

 So... oops, where's equation 4 gone? So by the cross terms I mean like abtimes bc and so on. So ab²c and we're gonna get a²bc and abc².Okay and then on the right we've got 9. Okay. Now with this bit you might notice that we can factorise that because we've got abc in each term, sothat's abc times a + b + c. Well we know a + b + c,that was 4, and we know abc is -2,

 so we've got 4 times -2,that bit's -8, and then times that 2 is -16,so like that whole thing is negative 16. If we shift that over we'll get 25, sowe know that all of that is 25. Okay and then another thing that we could dobecause we're looking for a to the fourth so I sort of thought of two waysthat we might be able to get a to the fourth. We could do equation 1 timesequation 3; that didn't work out too well. Well maybe it does, I just couldn'tsee a way to make it work out. The other way that you could do it is you cansquare equation 2 

because that's going to give you an a to the fourth and so on,so let's try that. So that will give you a⁴ + b⁴ + c⁴ and again2 times the cross terms, so a²c², b²c² and a²b², and that was10 so 10 squared is 100. Okay well now we've got that that is 25,so that whole thing is 50, we can shift that across, and we get that that is 50(because 100 - 50) so that is the answer to that one.

 Okay so I went back and had another go at substitution method and worked it out this time. So I don't wantto spend too long on this but let me just quickly go through what I did. So Itook that equation and solved it for c and then subbed it into equation 2, and played around with that and got to there, and then subbed it in to equation 3 as well, and that was annoying to expand that cubed but yeah you can do that. Some stuff cancels.And then what I did was instead of solving this for b which would be kindof like the standard way to do substitution. You can do that and then you get like... it's a quadratic in b and so it's like plus/minus square roots of stuff and that's where

 I gave up before because it was a bit too messy.But instead of doing that what I did this time was I decided to solve thisfor b² and then sub that in there and also in there and see what happens.And it turns out that a lot of stuff cancels and you just end up with thiscubic. Now if you just try a few values like often you'll get solutions for, you know,if you try a = 1 and a = -1 and 2 and -2 and so on,you might find some solutions. So I tried a = 2 and that worked; that gave 0

. Sowhat you can do then is divide that by a minus 2 because you know that that's a factor of it, so if what is equation you divide that by a minus 2 you get a² - 2a - 1so either you've got a = 2 or this equals 0, so solve that just using quadratic formula you would get a equals 1 plus or minus root 2. So because of the symmetry of the situation, where you've got like a and b and c are essentiallyinterchangeable, 

like if you swap a and b the system of equations are still like the same thing. That means that these solutions for a are also solutions for b and c, so it might be like a equals b... sorry a=2, b = 1 + root 2,and c = 1 - root 2. So then to work out a⁴ + b⁴ + c⁴, you just raise all those things to the power 4. So Idid that. If you expand those then a lot of stuff cancels, like you've got the+4√2 there and the -4√2 there. The stuff that's left after you've cancelled... you've got 1 and 12 and 4 that makes 17 and then you've also got 17 there and the 2⁴ is 16, so you add all those up and you get 50. what is equation

So that's another way that you can do it. The solutions from the booklet actually did it sort of a way that was in between those two,because what they did was they started out the same as what I was doing before, withlike elimination basically, but once they worked out abc and ab + bc + ac,they used that to find the coefficients of the cubic.