Hardest maths questions - solve equation involving cube roots

This was a question from the 1988 Australian Mathematics Competition senior division (for students around 17 years old) and only 3% got it right. It's a question about solving an equation involving cube and square roots, and is surprisingly linked with the history of complex numbers!

[tthtlc]

thank you...this question is very addictive: just cannot help NOT to switch to other webpage until I solve this.

[hvok]

Wow, the presentation of this was really good. I appreciate the sprinkling of math history, alongside a general algebraic sketch of the problem before calculation - great video.

[satyanarayanmohanty]

Another approach would be
Taking
∛(n+√(n²+8))=a
∛(n-√(n²+8))=b
a+b=8
ab=-2
Solving for a and b
a=4+3√2 and b=4-3√2
now a³= n+√(n²+8) and b³= n-√(n²+8)
n= (a³+ b³)/2=216+64=280.

But yours is easier one. Thanks for uploading great problems with brilliant explanations.

[anhquocnguyen]

I love your channel.
It is load with cool problem

[ndlwsr]

@THaWoM Jenni, Excellent combination of the hows and whys with an understanding of the history of this particular type of equation solving!

[quantumphase]

Hello, I'm preparing for the Australian maths olympiad (more like doing it for fun) but the solutions that I come up with are not as rigorous as the ones provided with the answers. Sometimes the solutions are so abstract that I cannot even understand them. Are there ways to sort of learn how to solve these types of questions from books, online resources or from a coach. I havnt come across a math coach.

[whatdoyouthink]

thanks dear teacher

god bless you.
please tell me the name of this formula.
because I wnat be familiar more with this formula.
thanks❤️

[ΩιλλιαμΛεε]

Easier approach would be to use the formula: from:korea

[rajeshmittal]

These questions will increase the efficiency of a person's mind
Thanks maam





I will request to god that your channel will become a great one in upcoming years

[jlmassir]

This problem can also be solved by direct comparison with Cardano's formula for x³+px+q=0. Taking -q/2=n and (p/3)³=8, It follows that 8 is a root of x³+6x-2n and therefore 8³+6·8-2n=0, so n=280.

[premparuchuri]

i am a IGCSE student and so the question for first time and the question was tricky very hard if can you explain igcse probelms
can you give number or email adderes for i have dout in it

[gidonkessler]

i don't know why i moved one cube root to the right...

good video

[omegamath]

Hmm! This is wrong way to find the answer.

so, this mathematical expression, (n+sqrt(n^2+8))^(1/3)+(n-sqrt(n^2+8))^(1/3), cannot be 8 at all.

[zalfanurafirst]

Thank u. I have similar question with different number

[rajeshmittal]

Can i also send you my challenging questions

If yes, then pls explain the procedure

[rajeshmittal]

Maam if you dont mind can i ask you from where do you belong???

[XHACKER]

I'm preparing for IIT JEE and I'm sure my method and answer is correct then can you mam please explain why my answer is not as same as your answer
---> n^1/3 + ((n^2+8)^1/2)^1/3 + (n)^1/3 - ((n^2 + 8)^1/2)^1/3 = 8
---> n^1/3 + (n)^1/3 + (2√2)^1/3 + (n)^1/3 - (n)^1/3 - (2√2)^1/3 = 8
---> 2n^1/3 = 8
--->. n^1/3 = 4
--->. n = 64

[achrafallali]

can you help me to find a solution for ; ∛(x+√x) -∛(x-√x) =2

[rajeshmittal]

Yeah, i got the correct answer
It is so simple
Only a minute question