Simplify Surds | NO CALCULATORS! | Purple Comet Math Meet 2017 Problem 17

[@nilsastrup8907]

WOOOOOOW. That's so beautiful... How a complicated expression like that can simplify so neatly is so nice!

[@ryanjagpal9457]

That is too complicated

[@randomjin9392]

Alternatively: set the whole thing to x, cube it. You will end up with √(26+15√3)-√(26-15√3) - 3x = x³ . Set the constant to y, square, you will end up with y² = 50, so y = 5√2. Thus the final equation is x³+3x-5√2 = 0 from where the x = √2 is an obvious solution and the factoring quadratic has to other roots.

[@haricharanbalasundaram3124]

Wow, that's a really neat method... I like this better than the one showcased!

[@ivarangquist9184]

Very cool indeed. However, please avoid using the word obvious like that.

[@randomjin9392]

Obviously, if the polynomial has a "simple" root, it must be some factor of the constant term. We have the constant as 5√2 so obvious choices would be ∓1, ∓5, ∓√2 and ∓5√2. Rational roots can be crossed out since obviously with the irrational constant they will never work. Original expression is obviously positive thus the only options left are √2 and 5√2. Now, 5√2 will obviously be too large when cubed meaning that the obvious only option left is √2. When trying it out it obviously works.

[@bjorneriksson2404]

Nice. I tried the guessing game. First, I guessed that the stuff inside the roots were perfect squares: (a√3 + b)², but that didn't work. Then that they were perfect cubes: (a√3 + b)³. Set the √3 terms = +/-15, the other terms = 26. That worked out, with a=+/-1, b=2 (from the √3 terms, you get a * (a² + b²) = +/-5, and for whole number solutions that 's obviously (+/-1) * 5). So, now we have √(2 +√3) - √(2 - √3). Square that expression and the answer is 2. Positive root = √2.

[@fredthelegend7673]

Nice 👍, I like this method, good thinking!

[@mcwulf25]

That's what I did 👍 I figured that sixth root meant we had a perfect square or a perfect cube inside.

[@SureshChoudhary-vh5hq]

If we want we can find the values of a and b a²+b²=4 (a+b)²-2ab=4 (a+b)²-2=4...(ab=1) (a+b)=√6 & (a-b)=√2 Therefore, a=(√6+√2)/2 & b=(√6-√2)/2 :-)

[@dennisdeng3045]

Lo and behold the sine and cosine value of 15 degrees!

[@jofx4051]

wtf 😂

[@adandap]

Very neat way to do this problem. I used a more brutish method. Write the expression as x and square, to get x^2 = -2 + cuberoot(+) + cuberoot(-). (+- signs inside surds). Then cube the result to get (x^2 + 2)^3 = 52 + 3 (cuberoot(+) + cuberoot(-)) = 52 + 3 (x^2 +2). So you get a polynomial equation x^6 + 6 x^4 + 9 x^2 -50 = 0. Probably easiest from there to set x^2 =t to make it a cubic t^3 + 6 t^2 + 9 t -50 = 0. It's easy to see that t=2 is a root and the other roots are complex. So x^2 = 2 and since x is clearly positive, x = sqrt(2).

[@MichaelRothwell1]

I got and solved the same sextic equation. The video solution was much neater than mine, but I was pleased to be able to solve it anyway!

[@debayuchakraborti1963]

BRUH PLS BRING BACK YOUR OLY MATH VIDEOS>>>PLSSSS !!!

[@letsthinkcritically]

This is a Maths Olympiad contest. One of the organisers of this contest is Prof. Titu Andreescu, and he was the head coach of the US IMO Team. I find problems in this contest very interesting.

[@debayuchakraborti1963]

​@@letsthinkcritically no like i was talking about the hard problems u brought previously like fe and number theory problems from the imo or usamo and the ideas behind these problems were brilliant and I really loved ur approach...I am really missing those kind of problems

[@jose4877]

@@letsthinkcritically Kind of funny how you still refer to Titu Andreescu as head coach of the USA IMO team. He hasn't been that in almost 2 decades. lmao.

[@chhabisarkar9057]

@@jose4877 whatever lol but someone please request this guy for bringing those good ol olympiad NT problems

[@particleonazock2246]

1:45 Reminds me of the Sophie Germain factorisation

[@kfho9783]

Thank you teacher 😊

[@gabrielsantana3861]

I'm from Brazil, love your videos, thanks!

[@mariogermano8887]

O sotaque desse cara é chique demais slc

[@abhisheksahgamerz4138]

Is any Indian is here with me 🖐🏻 Btw very tough question

[@pksingh9584]

Ya

[@mcwulf25]

(Set a + b√3)^3 = 26 + 15√3 and we find a=2, b=1. So the terms in the sixth roots are (2+√3)^3 and (2-√3)^3. Take the sixth roots and those same terms will now be y = √(2+√3) - √(2-√3). Square both sides and we find the cube roots disappear and y^2 = 2. 👍

[@tyx7650]

Love your vids

[@jose4877]

Nice trickery.

[@zazamedaouar5955]

Hi I am from Algierie and I love math &YOUR vidio also 😍🤓🤓🤓🤓🤓🧠🧠🧠🎒🎒

[@txikitofandango]

I did it by rewriting the problem as a-b and expanding (a-b)^6. Then, a^6 + b^6 = 52 and ab=1. By repeatedly using these facts, you can get the original binomial expansion down in terms of x=(a-b)^2, and then all you have to do is solve the cubic x^3 + 6x^2 - 39x + 46 = 0.

[@242math]

this was complicated but you easily solved it, great job

[@shijinnambiar9]

This question can also be done as 26 +15√3 is a perfect cube root of 2+√3 and same as 26-15√3 of 2-√3.

[@MichaelRothwell1]

Very neat. I thought of trying that approach but ended up doing it a different way, more similar to the video.

[@engjayah]

Alternatively: square the whole expression (y) and set x = (26+15sqr3)^1/3; we end up with y^2 = x + Conj(x) - 2 ; now set x + Conj(x) = m; evaluate m^3; from which we end up with m^3 - 3m - 52 = 0 and m = 4; back substituting y^2 = 4-2 and we get y = sqrt 2

[@tonyhaddad1394]

Wowwwwwwwwww man that onctly that is the beaty of math always share with us this amazing problems !!!

[@riothegoldenretriever6909]

really loved it man

[@ManuelRuiz-xi7bt]

I just used a brute method. With X = the expression to evaluate. General cube: (A+B√3)³ = A(A²+9B²) + 3B(A²+B²)√3 = 26+15√3 = (2+√3)³. Similar for opposite B. So X = √(2+√3) - √(|2-√3|) = √(2+√3) - √(2-√3) General square: (C+D√3)² = (C²+3D²) + (2CD)√3 = 2+√3 = [(1+√3)/√2]² . Similar for opposite D. So X = (1+√3)/√2 - |1-√3|/√2 = (1+√3)/√2 - (-1+√3)/√2 = 2/√2 = √2.

[@abhitomar320]

Great

[@echandler]

Note that (26+15√3)(26-15√3) = 26²-3(15²) = 26²-3³5² = 26²-27(25) = 26²-(26+1)(26-1) = 26²-(26²-1) = 1

[@rohitrajesh2933]

(26+15√3)=(2+√3)³=((√3+1)⁶/8) (26-15√3)=(2-√3)³=((√3-1)⁶/8) After taking sixth root and subtracting, ((√3+1)/√2)-((√3-1)/√2) 2/√2 √2

[@leecherlarry]

*(26 + 15 Sqrt[3])^(1/6) - (26 - 15 Sqrt[3])^(1/6) // FullSimplify*

[@shijinnambiar9]

In india these type of questions are studied in class 9th

[@ryanjagpal9457]

Where did you get 2a^2b^2 and -3a^2b^2 Why would x also be equal to (x+2)^2 + 9? Why did he do (a-b)^2?

[@HassanLakiss]

Thank you. Is this the only way to do this question?

[@tester-h4y]

can u do a vid of the volume of a twisted cube(the bottom of the cube horizontally turn 90 deg) with side = s? i want to confirm the answer i got :p (i got (2s^2)/3 by calculus btw)

[@skyhipeofflical]

Level🔥😩

[@waqaramin3400]

Sir Please explain this 4^3+3(4)-+2=0.........

[@giuseppemalaguti435]

6rad3/5rad2

[@Mathstoon]

wow amazing #mathtutorial

[@indrakumaryadav021]

Write question completely 🔥🔥🎉

[@aashsyed1277]

So good....

[@acident4939]

This is something you’ll learn how to simplify in common 9th grade math in my country.

[@officialyoutubecucumber8797]

Where are you from??

[@ramaprasadghosh717]

ab = 1 a^6 + b^6 = 52 (a^2+b^2)^3 - 3(a^2+b^2*)a^2*b^2 =52 so (a^2+b^2)^3 - 3(a^4+b^2) = 52 again remaining part is same

[@gervasiociampin2062]

Why is ab=1? Please

[@MichaelRothwell1]

@@gervasiociampin2062 if you multiply what's inside the 6th roots you get (26+15√3)(26-15√3)=676-225×3=1