Hardest maths questions - find a^4 + b^4 + c^4 

We can also go by polynomial approach:- Let us assume that a,b and c are zeroes of some cubic polynomial f(x). So we have f(x) =(x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now our task is to simply find those coefficients. a+b+c = 4 2(ab+bc+ca) =(a+b+c)² - a² - b² - c² = 4² - 10= 6 Hence, ab + bc + ca= 3 Using identity: a³+b³+c³ - 3abc =(a+b+c)(a²+b²+c²-ab-bc-ca) We get abc = -2 =>f(x) = x³ - 4x²+3x +2 Since 'a' is a zero of f(x), a³ - 4a²+3a+2= 0 After multiplying both sides of this equation by 'a' and a little bit of rearranging, we get: a⁴ =4 a³ +-3a² -2a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = 4(a³+b³+c³)-3(a²+b²+c²) -2(a+b+c)=88-30-8=50.

Let f(n) = a^n + b^n + c^n, then f(n) = f(1)•f(n-1) - (ab + bc + ca)•f(n-2) + abc• f(n-3). In this case, abc = -2, ab + bc + ca = 3.

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.

we had to solve this equation on our math wizards competition and thank god i've watched this video!

a+b+c=4 a²+b²+c²=10 a³+b³+c³=22 a⁴+b⁴+c⁴=? Lets consider them as EQU 1,EQU 2and EQU 3 Then subtract EQU 2-1 10-4=6 While subtraction of EQU 3-2 22-10=12 As it is coming in 6 tables then the next pattern of it is 18 Then 22+18=40 That's it...

Thanks you mam I am just 15 year's boy I am in 10 th class so this problem is most important to my life

Using Newton-Raphson you can solve for a,b,c in about 5 iterations. Did the competition allow the use of spreadsheets?

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

You could also have found it without finding the exact value of a, b, c. Just by squaring the new equations.

I got a cubic eqn. x^3-4x^2+3x+2=0 So one value is 2 from simple observation Next dividing by x-2 we get x^2-2x-1=0 It was smooth sailing after that

Ma’am, you’re absolute genius♥️

One thing that irks me about these maths "challenge" questions is it seems to be more that you just have to get lucky and go down the right path rather than having any actual maths ability. The maths I do on my Uni physics course may be harder, but it is at least fair, hence I can do it while having no idea how to solve this. You seemingly just shove the equations together and hope something works. There may be a way to solve it more simply, but without being taught that method beforehand you're not going to figure it out during the exam, turning the question into trial and error. Mostly error.

The expression a^4 + b^4 + c^4 represents the sum of the fourth powers of three variables: a, b, and c. To simplify this expression, we can leave it as it is since there are no like terms to combine. However, if you are looking for a factorization or a specific form, we can use the identity known as the "Sophie Germain identity" to rewrite it: a^4 + b^4 + c^4 = (a^2)^2 + 2(a^2)(b^2) + (b^2)^2 - 2(a^2)(b^2) + (c^2)^2 - 2(b^2)(c^2) + 2(a^2)(b^2) + 2(b^2)(c^2) - 2(a^2)(c^2) = (a^2 + b^2)^2 + (b^2 + c^2)^2 + (c^2 + a^2)^2 - 2(a^2)(b^2) - 2(b^2)(c^2) - 2(a^2)(c^2) So, the expression a^4 + b^4 + c^4 can be written as the sum of squares of various combinations of a, b, and c, with some additional terms involving products of squares.

Dude your process is cool but I've quite simpler method how do i share that with u???

Thank you for watching🙏

Very complicated to solved but correct, I can solve this problem in very easy way

It seems that I solved it in the way it's done in the book. Actually, I was challenged to solve this problem by the guy from my university, and he said there's a formula, knowing which you can solve it in 2-3 minutes (these words got me really interested). After getting through it I was trying to find one for looong hours. Although I failed, it was a lot of fun :)

Ma'am it is too easy i solve it in less then 1 minute. My way of solving is completely different than your's. And I want to show you its simplest way.

Ma'am I solved it and it was actually good to. Which standard is this of anyway?

So if I may ask, where can we find the right answer? As I got my answer but now I would like to know the real answer haha 😄

This question was asked in IIT JEE ( Indian institute of technology joint entrance examination) (2016) in India 🇮🇳 and more than 94% of students answered correct

√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.

If 4-10 is 6 then a2+b2+c3 is 6 more than four then in a3+b3+c3 is 22 than 22-10 is 12 is double of 6 and a4+b4+c4 is 12 multiply 2 which is 24 so 22+24 is 46 so A4+b4+c4 is 46

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

Easy question I had solved before seeing solution with very simple way by using only algebraic identity 😄😁😁 and I am in class10 student in 2021-22

Why am I watching this? I’m literally 9 years old

I SOLVED IT I FEEL SO PROUD OF MYSELF

So good

This is the appropriate formula a^4+b^4+c^4 = (a^2+b^2+c^2)^2 - 2(ab+bc+ca)^2 + 4abc(a+b+c) ------------(1) Given a+b+c = 4 a^2+b^2+c^2 = 10 We know (a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) Using the above we can determine (ab+bc+ca) = 3 Given a^3+b^3+c^3 = 22 We know a^3+b^3+c^3 = (a+b+c) (a^2+b^2+c^2 - ab-bc-ca) + 3abc Using the above we determine abc = -2 Substitute the values in the formula (1) a^4+b^4+c^4 = (10)^2 - 2 (3)^2 + 4 (-2) (4) = 100-18-32 = 50

Formula=answer of upper question×2+2

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

Didn't understand much but you are cute and your voice and accent is classy.

Can't believe the answer was that obvious and I could not see it!!!

Is it 46?

Isn’t the answer 40? 10-4=6. 22-10=12. And since it’s six’s table the next number would be 18 so, 22+18=40

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 think the answer was 46 1st answer =4 2nd answer=(4+6)=10 3rd answer=(4+6+6)=22 so the 4th answer is = (22+6+6)=46 i am just 11 years old

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

Easy way find the difference between 4 and 10 which is 6,Difference between 10 and 22 which is 12 so times it by 2 =24.Add 24 onto 22 which makes 46.Easy.

I write on a board

Symmetric functiions Newton's formula for sum of the powers and Vieta formulas

as a 6th grader I am NOT READY for this type of math-

The answer is 42.

Why did you erase 1 from 16 in first question...

(a+b+c) (a²+b²+c²) may i know why is 40 answers i didn't get why 40 how did his make 40? exp 20+20=40 it's clear for me the answer of 20+20 is equal 40, my question is where do i know the answer is 40?

Nice math question❤❤❤😢😢😢

Your mom is looking like you ❤😊

I understand very well

Baba Bööööey

can u do some from the junior division

Excellent... Love ❤️

Mam what is the answer of last one

I like it

what is the answer of this

1. a + b + c = 4 ... (1) 2. a² + b² + c² = 10 ... (2) 3. a³ + b³ + c³ = 22 ... (3) We can start by finding the value of a⁴ + b⁴ + c⁴ using the following identity: a⁴ + b⁴ + c⁴ = (a² + b² + c²)² - 2(a²b² + b²c² + c²a²) Using equation (2), we know that a² + b² + c² = 10, so: a⁴ + b⁴ + c⁴ = 10² - 2(a²b² + b²c² + c²a²) Now, let's find the value of a²b² + b²c² + c²a²: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) Using equation (1), we know that a + b + c = 4, so: 16 = a² + b² + c² + 2(ab + bc + ca) Substituting a² + b² + c² = 10 from equation (2), we get: 16 = 10 + 2(ab + bc + ca) Simplifying, we get: ab + bc + ca = 3 Now, let's find the value of a²b² + b²c² + c²a²: a²b² + b²c² + c²a² = (ab)² + (bc)² + (ca)² Using the fact that (ab)² + (bc)² + (ca)² ≤ (a² + b² + c²)(ab + bc + ca), we get: a²b² + b²c² + c²a² ≤ (10)(3) a²b² + b²c² + c²a² ≤ 30 Now, let's find the value of a⁴ + b⁴ + c⁴: a⁴ + b⁴ + c⁴ = 10² - 2(a²b² + b²c² + c²a²) a⁴ + b⁴ + c⁴ ≥ 100 - 2(30) a⁴ + b⁴ + c⁴ ≥ 40 So, the minimum value of a⁴ + b⁴ + c⁴ is 40. Therefore, the answer is: a⁴ + b⁴ + c⁴ ≥ 40

i really appreciate your brain . I solved some more similar question by those methods By the way love from Nepal

Yeah I don’t really know why I search this even though I’m still grade 3

Newton girad identify always help

what was the answer ?

I Am online from India 👍

Yeah I got the correct answer It is quite simple But it took lots of brain

Answer is 297 🥰

Nice 🙏 from 🇮🇳 INDIA

What the fuck is the awnser

It's a fudge!

42 is it correct?

Challenge: solve for a , b and c.

lovely

great 👍👌

50 is the answer?

I was able to solve it DAB!

इसी लिए देश में खान सर जैसे टीचर की जरूरत है.

Why doesn't this work? : a+b+c=4 a^3+b^3+c^3=22 (a+b+c)(a^3+b^3+c^3)=a^4+b^4+c^4=x 4*22=88=x

Mam the answer is 40

a⁴+b⁴+c⁴=x (a²)²+(b²)²+(c²)²=x (a²+b²+c²)²=x 10²=100

just think about it they added 2 more for eVeRy

I solved it, I can't believe it!

The answer us 244

i solved this question in an entirely different way and got the answer

Mam super u r a genius 🥰

It's answer is 80

Smart

Iska answer kya hai

your given was wrong

Mam please give me short ans

Ans 40

48

Questions samjha nhi

Is it 40???

If 4-10

simetry polinom ..newton

I solved this problem in 20mins by just saying one word.....Gauranga....

I m 10 years old :)

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

Me: a+b+c = 4 a^+b^+c^ = 10 ( more 6) a^3+b^3+c^3= 22 ( more 12) So... a^4+b^4+c^4 = 22+24 = 46 rigth? Is not........!!! Omg wtf

Thank you miss

is the answer 50

This is very ez

Answer is 40

I got it right. I am 7*pi years old.