# Hardest maths questions – find a^4 + b^4 + c^4

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.

Here is what I found (before checking your solution)

(First I made an error, because I wrote a²+b²+c² = 16 -> anwers a^4+b^4+c^4 = 32)

(A) : a + b + c = 4

(B) : a^2 + b^2 + c^2 = 10

(C) : a^3 + b^3 + c^3 = 22

(D) : a^4 + b^4 + c^4 = ?

(A)*(A) = 16

(a+b+c)(a+b+c) = a^2 +b^2 +c^2 + 2(ab+ac+bc) = 16

use (B) a^2+b^2+c^2 = 10

then

ab+ac+bc = 3 (E)

(E)*(E) = 9

(ab+ac+bc)(ab+ac+bc)

= a^2 b^2+a^2 c2+b^2 c^2 +2( a^2 bc + ab^2 c + abc^2)

= a^2 b^2 +a^2 c^2 +b^2 c^2 + 2abc(a+b+c)

use (A) a + b + c = 4

= a^2 b^2 +a^2 c^2 +b^2 c^2 + 8abc = 9

then

a^2 b^2 +a^2 c^2 +b^2 c^2 = 9-8abc (F)

(A)(B) = 10*4 = 40

(a^2+b^2+c^2)(a+b+c)

= a^3+b^3+c^3 + ba^2 + ca^2 + ab^2 + cb^2 + ac^2 + bc^2

= a^3+b^3+c^3 + a(ab + ac) + b(ab + bc) + c(ac+b)

use (E) ab+ac+bc=3 and (C)

a^3 + b^3 + c^3 = 22

= 22 +3(a+b+c) – 3abc = 40

= 22 +12 – 3abc = 40

then

abc = -(40-22-12)/3 = -6/3 = -2 (G)

(B)(B) = 10^2

= 100

(a^2+b^2+c^2)(a^2+b^2+c^2)

= a^4+b^4+c^4 + 2(a^2 b^2 + a^2 c^2 + b^2 c^2)

use (F) a^2 b^2 +a^2 c^2 +b^2 c^2 = 9-8abc

= a^4+b^4+c^4 + 2(9-8abc)

= a^4+b^4+c^4 +18 -16 abc

use (G) abc = -2

= a^4+b^4+c^4 + 18 -16(-2) = 100

then

a^4+b^4+c^4 = 100 – 32 – 18

a^4+b^4+c^4 = 100 – 50

a^4+b^4+c^4 = 50

=> (D) a^4+b^4+c^4 = 50

I SOLVED IT I FEEL SO PROUD OF MYSELF

Yeah I got the correct answer

It is quite simple

But it took lots of brain

I was able to solve it DAB!

not hard

Thank you I solved it myself ……quite easy ….as per indian curriculum……we assume integration to be more difficult

But very nice and lucid explanation ……

From India

can u do some from the junior division

you teach very well mam

I am actually proud of myself! I was on the right track on the answer but got stuck using substitution! Then I used elimination and got the answer really easily! I am only 11! I love those challenging maths videos! Please make more! I don’t really care what age it is I like a challenge!

44 years old and feel so left behind. This is a high school math problem? Dont do drugs, ditch your friends and stay in school.

I solved it, I can't believe it!

50.

I got the answer 147…… I may have made an error…..

isn't quite a hard question. at least you know the only way to go is to use substition. given enough time by brutal force, you will definitely get it

i really appreciate your brain . I solved some more similar question by those methods

By the way love from Nepal

In india we are taught an easy method to solve the problem. (These are level one questions in our olympiads)The real question is can this be generalized for n exponents

Ie give a^n-1 + b^n-1 + c^n-1 and all other sums below it and solve for a^n + b^n +c^n

Oh my God I can't believe I got it right the Miracles that I'm only 13

A way to get close to the answer which I found quite genius is a equals two vehicles one and sequels when you add that all up on the very first equation then you do the work on the second equation to find out that it only equals 6 then you add it with the first equation answer and then for the third equation what you do is that you add the third equation with the fourth equation to get 22 send to get a for what you have to do is do the exact same thing what you've been doing for the other equations which is you know multiply them by the two then by the one and then by the one and then what you have to do next is you actually have to do next is at it with the next equation which is 5 and then when if you add that up you get 52 then you subtract that by 18 then you add it + 10 + 6 and you have 50

a²+b²+c²=16..x²+y²+z²=25..ax+by+cz=20..then what is the value of (a+b+c)/(x+y+z)..?

I got it right and im only -2 years old.

√2 + y+ z = 8

x +√y +z =12

x +y + √z =14

Find x,y,z.

Please send me its solution (shortest method).

I hope you will answer me.

Quite easy to solve isn't

I can solve this problem only one minute

my head is rotating with earth.lol.keep posting more.Wish u become as popular as paresh from mind your decisions.love from india