Okay this question is from the 2006 competition and only 1% of students got
this one right, although this one isn't multiple-choice so that does make it
harder. All right, if you haven't tried this one before then pause the video
here and have a go. 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 "b squared"s and maybe that's going to be
able to be combined with equation two somehow. 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 basically 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 gonna be 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 we just 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. 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 was helpful, 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 out everything 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 b²c plus another abc
plus another abc from there, plus bc² and ac² 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 what we had here, right? Because we've got the
a²b which was there and a²c which was there and so on, and we
know that all of that is 18, so we'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 get negative 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 4 squared, so that squared. You're gonna
get like the square of each one and then plus 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 and then 2 times the cross terms.
So... oops, where's equation 4 gone?
So by the cross terms I mean like ab times 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, so that'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, so we know that all of that is 25. Okay and
then another thing that we could do because we're looking for a to the
fourth so I sort of thought of two ways that we might be able to get a to the
fourth. We could do equation 1 times equation 3; that didn't work out too
well. Well maybe it does, I just couldn't see a way to make it work out. The other
way that you could do it is you can square 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 again 2 times the cross terms, so a²c²,
b²c² and a²b², and that was 10 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 want to spend too long on this but let me
just quickly go through what I did. So I took 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 kind of 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 this for 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 this cubic. 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. So what you can do then is divide that by a
minus 2 because you know that that's a factor of it, so if you divide that by a
minus 2 you get a² - 2a - 1 so 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 essentially interchangeable, 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 I did 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. 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, with like elimination basically, but once they
worked out abc and ab + bc + ac, they used that to find the coefficients
of the cubic. So they went like straight from that to this point, and then did the
same thing from there. So there's like three different methods so far that I
know of that you can use to solve this. Okay let me know if you have any
questions and stay tuned for the next video. You might also like to check out
the inheritance question that I did for the junior students because that was a
senior question as well and only 8% of the seniors got that one right.