The greatest trick humanity ever learned from mathematics - If you don't know some value, call it x and proceed.

Forget the math, I want to learn that marker-switching trick.

If you look just under the input box after he hits equals there is a blue box that contains the text "Assuming the principal root | Use the real-valued root instead." Click on the option for the real-valued root and it gives you 2. I do not know why Wolfram Alpha defaults to the principle root. But it will calculate the real value correctly. The same is true by the way when you use Mathematica by the way. However, I am not sure how to get Mathematica to return the real-valued root.

When you get an integer as a root, that makes you reeeeaaaally happy.

At 13:36 you can see that wolfram alpha is assuming the principal root is being used. If you click the option to use the real-valued root instead you will get the answer of 2.

On Wolfram Alpha, you're showing the principle root. If you click "Use the Real Root instead," you'll see your expected answer of 2. That being said, I agree that it's bad to lean on Wolfram Alpha and Mathematica too much. But they certainly are nice tools when used correctly!

Whoever came up with this problem is a genius

Russians do like this: Cuberoot(7-5sqrt(2))=cuberoot(1-sqrt(2))^3=1-sqrt(2). Cuberoot(7+5sqrt(2))=1+sqrt(2). Finally, (1+sqrt(2)) + (1-sqrt(2))=2.

Your marker switching skills are deft af.

I feel like I disappointed him every time I don't pause to try 😅

2:02 the way you swap between the black pen and red pen is so smooth, Jesus.

This equation literally blew my mind.

The marker switch game is strong in this one.

I always enjoy when some complicated sq roots, cube roots etc. end up equaling some integer value :)

I believe wolfram alpha just shows you one of the correct answer, one of the complex solutions, because the cubic root of something actually leads to three results, not just one. You just can't take it as a real number for granted.

I tried it before watching the video, and I got the same thing by essentially the same method. I approached the end with some slightly different phrasing, though. I did work it down to 14-3x=x^3. But I *just* moved the 3x to the other side, giving me 14=x^3+3x. I factored out the common x: 14=x(x^2+3). Knowing that 14 is composite (and seeing immediately that x=1 didn't work), I factored it into 2 and 7 (conveniently, its full prime factorization) and started testing. It turned out that 2 worked, as 2 * (2^2+3) = 2*(4+3) = 2*7=14.

Impressive black pen red pen powers at 2:00

2 7+5sqrt2 is 1+3sqrt2+6+2sqrt2 which is (1+sqrt2)^3 which means a=1+sqrt2 We do same for b and b=1-sqrt2 a+b=2

In WolframAlpha you have to use the Cbrt function which is the real-valued root: Cbrt[7+sqrt(50)]+Cbrt[7-sqrt(50)]. And the result is "2"

The reason you get the incorrect answer is because you miss cube root of -1. at 7:30 you cannot take (-1)^1/3 outside the cube root. cube root of -1 has 3 roots so you should get three different cubic equations for X. That's why wolramalpha does not give 2 as the result

at the top it says: Assuming the principal root | Use the real‐valued root instead click the blue bit.. it shows 2! so dont complain about wolfram alpha result untilk you READ THE RESULT

What really impresses me is how seamlessly you can switch between both pens. Amazing.

Lovely algebra. This sort of thing doesn't seem to come up much, but it's still great to know how to work it out, and your explanation was crystal clear. Obviously you have to take care, with x, when there's multiple roots, but only one is a valid solution.

This video really has given me some confidence in my mathematics adventures, I’m a Bio/Chem Major and have always loved math but haven’t touched it in awhile and I feel as if I could’ve completed this problem which was seemingly hard to begin with. Thanks bud.

click on "Use the real‐valued root instead"

We need a video on how to switch markers like that, looks simple yet amazing.

But if you enter this: cbrt(7-sqrt(50)) + cbrt(7+sqrt(50)) The result is 2.

Notice that 7+sqrt(50) can be also written as 7+5*sqrt(2) which is a formula of a cube 1^3+3*(1^2)*sqrt(2)+3*1*(sqrt(2)^2)+sqrt(2)^3 and that is simply (1+sqrt(2))^3. Also 7-sqrt(50) is (1-sqrt(2))^3. Further-easier. (1+sqrt (2))^3^(1/3)+1(-sqrt(2)^3^(1/3)=1+sqrt(2)+1-sqrt(2)=2. Easy! But I can notice that your solution more universal if take another numbers. Great job!

Let the first term be "a" and the second one" b" and x=a+b. Notice that ab=-1 and a^3+b^3=14. Then a^2+b^2 =2+(a+b)^2 =2+x^2 And a^3+b^3=(a+b)(a^2-ab+b^2) =>14=x(3+x^2) And we then get the equation: x^3+3x-14=0

Its much easier if you write it like this: (a+b)^3 = a^3 + b^3 + 3ab(a+b) And there you can substitute the x: = a^3 + b^3 + 3abx

nobody: blackpenredpen: believe in the math

This is the first time that I've solved alone one of math problems after watching four or five of your other ones! and I'm happy with that . thank you for your intresting work and content!

If you type ∛(7+√50)+∛(7-√50) into WolframAlpha it gives the answer 2 If you type (7+√50)^(1/3)+(7-√50)^(1^3) it doesn't . ∛(7+√50)+∛(7-√50) uses the real root, (7+√50)^(1/3)+(7-√50)^(1^3) uses the principle root

When I'm using a bunch of completely different irrational numbers based on numbers that have no common factors and wind up with a rational number answer I look for where I messed up.

Thank you for your videos and your enthusiastic presentation, which I find very interesting. Usually you proceed at a blistering high speed, which is exciting and wonderful. This one looses momentum when you cube x numerically. If you do it symbolically, it would be faster and more transparent: (A+B)^3 = A^3 + B^3 + 3.A.B.(A+B) before you substitute numbers.

I got almost the same results, however I do have a doubt and it is that since cbrt(-1) has 2 imaginary roots too the final equations I got were S^3 + 3S - 14 = 0 ; S^3 - 3((1 + i sqrt(3))/2) - 14 = 0 ; S^3 - 3((1 - i sqrt(3))/2) - 14 = 0 I will say however that I am not very experienced with complex numbers so I may be wrong

WolframAlpha was correct, it is just a computer, though, and cannot know which solution you were looking for. That is why it asked you, whether you want to use the principal root, or the real valued cube root. You missed that, that's why it came with a complex solution, as people before me already pointed out.

Did you try clicking the link to use the real valued root, instead? I bet that's where your 2 went. Love the videos!

Everything is better if x^(1/3) is changed to cbrt(x)

can we just appreciate how easy he makes the pen swapping seem

I didn't even use wolfram alpha. I just used windows calculator and got 1.999999999... as the answer and I'm like ok, I guess it's 2 then.

13:38 Isn't there some hyperlinked text at Wolfram Alpha saying *use the real-valued root instead* - or am I missing the point?

calculator: Am I a joke to you?

why not call 7 + sqrt ( 50 ) = a^3 and 7 - sqrt ( 50 ) = b^3 then it becomes cbrt ( a^3 ) + cbrt ( b^3 ) = a + b Now call a + b = x ( a + b )^3 = x^3 a^3 + b^3 + 3ab ( a + b ) = x^3 14 + 3ab *x = x^3 to find ab we do : a^3 * b^3 = ( ab )^3 == > 49 - 50 = (ab)^3 == > ab = - 1 so as a result x^3 + 3x - 14 = 0 and x = 2 comes out.

I think it would be more efficient to modify binomial formula to A³ + 3AB(A+B) + B³ and then do the substitution. Also, as far as my experience goes, problems of this kind are constructed so that a perfect cube (square) is under a cube (square) root. In this case, the first try (1+√2)³=1+3√2 + 3*2 + 2√2 = 7+5√2 gives the answer right away.

Turns out wolframalpha wasn't wrong so I keep trusting it more than anything.

there is an easier method; assume k=expression. then substitute t as 7+root(50) then, 7-root(50)=-1/t therefore, k=(t^1/3)-{1/(t^1/3)} now cube both sides using (a-b)^3=a^3-b^3-3ab(a-b) a^3+b^3 becomes 14 ab becomes 1 and (a-b) becomes k again. solve the cubic to get the answer.

That's amazing. A small critique, when you referenced Pascal's triangle, you didn't mention that it is be specifically the third row. Making that association might have made it click for people that didn't quite grab the concept yet.

In my complex analyss classes (and a few books on the subject I use ∛ to denote the principal cube root only and reserve the superscript notation z^(1/3) to denote the multivalued power function. So, by that notation, Wolfram Alpha's answer is correct.

Thanks, KZfaq ads, for thinking that every math video on KZfaq is over a decade old and completely unhelpful... Great video, by the way!

cuberoot (7 +√50) + cuberoot (7 -√50), use this expression you'll get 2 in Wolframalpha. It is always good to solve math problems rather than depending on software tools.

Mr Blackpenredpen - a lovely problem but I’m afraid your solution to it is not. I paused the video and very quickly came up with the following. The key is that we are given there is an elegant solution so we know the expression can be simplified. ∛(7 + √50) + ∛(7 - √50) Start by realising that ∛(7 + √50) + ∛(7 - √50) = ∛(7 + 5√2) + ∛(7 - 5√2) Now the only way this will simplify is if the expression inside each of the radicals is a perfect cube. Let’s try and look for this. Can (7 + 5√2) be expressed as (α + k√2)³ ? You give us the expansion (a + b)³ = a³ + 3a²b + 3ab² + b³ Now in (α + β √2 )³ , we must have α = 1. If α ≥ 2, then we could not have 7 in the radical because 2³=8 Now we need β so that (1 + β√2)³ ≡ (7 + 5√2) By comparing this with (a + b)³ = a³ + 3a²b + 3ab² + b³ We can easily find that β = 1 So (7 + 5√2) = (1 + √2)³ similarly (7 - 5√2) = (1 - √2)³ So ∛(7 + √50) + ∛(7 - √50) = ∛(7 + 5√2) + ∛(7 - 5√2) = ∛(1 + √2)³ + ∛(1 - √2)³ = 1 + √2 + 1 - √2 = 2 No need to cover the white board in lots of horrible expressions and no need to try and guess the solutions to a cubic.

I like how complex mathematical formulas equal something so pure and simple

Something interesting I noticed: What you solved for in this video is a value of 7 for x in this equation: y=cbrt(x+sqrt(x^2+1))+cbrt(x-sqrt(x^2+1)) If you solve for x, you get x=(y^3+3y)/2 We can then prove that if y is an integer, x will also be an integer. To do this, we just need to prove that y^3+3y is even for all integer values of y, since an even divided by 2 is always an integer. This will be true when y^3 and 3y are either both odd or both even, since they must add up to an even number. y^3=y*y*y, and since an odd times an odd is odd and an even times an even is even, y^3 will be odd if y is odd and even if y is even. This is also true for 3y, since an even times an odd is even and an odd times an odd is odd. Thus, we can say that y^3+3y is even, and that by extension x is an integer for all integer values of y. Therefore, there are infinitely many integer solutions to the equation at the top of the post.

13:35 "Assuming the principal root | Use the real-valued root instead[.]"

50=49+1=(-7)^2+1^3. => x^3+3x-14=0 14 is the product of the roots, so if x is an integer, x \elm \pm (1,2,7) also has to be positive since the sign of 14 is negative. Then you. An solve by inspection (x=2)

The definition of cube root is different in Complex Analysis than the definition in Real Analysis. In Real Analysis, the answer will be 2 but in Complex Analysis the answer will be what Wolfram Alpha gives. Wolfram Alpha always uses Complex Analysis as a default. Maybe you should have mentioned whether you were working in Real Analysis or Complex Analysis or your definition of the cube roots. In Real Analysis, you can only have one cube root but in Complex Analysis the most sensible cube root is complex.

Don't believe in Wolfram Alpfa. Believe in blackpenredpen who believes in you.

"Use the real‐valued root instead"

You can solve it in an even better way Just take a look ³√a +³√b =X ³√a +³√b +(-x)=0 Now use the law that if x+y+z=0 then, x³+y³+z³=3xyz hope it helps @blackpenredpen

Wolframalpha is just "too wise". You have to input the formula as cbrt(7+sqrt(50))+cbrt(7-sqrt(50)) to get the real solution. Note that the issue here is that the expression x^(1/n) is not a function of x, and wolframalpha decides to report only one of the possible solutions of the equation y = x^(1/n) (the principal value).

I have a math exam on Thursday and something like this will probably be on it lol. Good video.

you should've told that 7 + (50)^(1/2) = (1 + (2)^(1/2))^3 7 - (50)^(1/2) = (1 - (2)^(1/2))^3 to not make confuses before you go into algebra trick or after.

I found a shortcut with almost no calculations: let's note a=(7+sqrt(50))^(1/3) and b=(7-sqrt(50))^(1/3). We want to find their sum s=a+b. Notice that a^3+b^3=14 (evident) Also we can easily calculate the product ab=[(7-sqrt(50))(7+sqrt(50)]^(1/3)=[49-50]^(1/3) =-1. Let us now consider the polynom P(X)=(X-a) (X-b). By developing it, P(X)=X^2- sX- 1. In fact, i replaced the product ab by its value which is -1. 0=P(a)+P(b)=(a^2+b^2)-s(a+b)-2 Which means that a^2+b^2=s^2+2 Also, 0=a P(a)+b P(b)= (a^3+b^3)-s(a^2+b^2)-(a+b) But remind that a^3+b^3=14 Then, 0=14-s(a^2+b^2)-s=14-s(s^2+2)-s Then s^3+3s -14=0 * At this point, he made a kind of 'trick' ( to not say big error) : in fact, we can verify that s:=2 does satisfy the equality * but eventually there 3 possible solutions for it and anyone of then could be the searched sum of a and b. His argument that the 2 other solutions are imaginary has to be proved . To do so, we manage to factorize s^3+3s-14 by (s-2) using the identification method and we get that s^3+s-14=(s-2)(s^2+2s+7) Thus the 2 remaining solutions are the zeros of s^2+2s+7 which is evidently having imaginary zeros. Consequently, we have a+b = 2 :) Notice that this method can be generalized : suppose that a=(y+sqrt(y^2+1))^(1/3)and b=(y-sqrt(y^2+1))^(1/3) such that y is an entire number (in the video he took y=7) then using my method you will found that their sum s=a+b is a solution of s^3+s-2y = 0. For example, if y=1 --> s= 1 if y=7 --> s= 2 (case of the video) if y=15 --> s= 3 if y=34 --> s=4 if y=75 --> s= 5 etc..... If you have any remark/question about my method please react to my comment :)

Since you use ^(1/3) instead to obtain the cubic root, so the principal root is returned. Try out (-1)^(1/3) and you will obtain 0.5 + 0.866... i (1/2+(√3̅/2) i) instead of -1. In polar representation of complex plane, -1 is represented as (1,180°(π)), so wolfram alpha should return (1,60°(π/3)) by default instead of (1,180°(π)) or (1,300°(5π/3))), the 180°(π) returns the result of 60°(π/3), different from that a positive real number which is 0°(0), returning 0°(0). In order to obtain the cubic root, choose real‐valued root or CubeRoot(-1) or cbrt(-1) instead

I like your channel. Really good stuff. You explain pretty well. Glad I found your channel. Thanks for all the videos.

(1+ √a)^3 = 1+3a + (3+a)√a similarly; (1-√a)^3 = 1+3a - (3+a)√a. So if you have an expression A +/- B√a; test special case: a+3 = B and; 1+ 3a = A So we have: 1 + 5√2 + 1 - 5√2 = 2

Thanks for the task, I didn't solve it completely, but had an idea to make like you did, that the new expression after ^3 starts too look quite same as the initial one. Really enjoyed the solving part, idk why, but for me it it's beautiful! Thanks!

let's assume that a€Z and b is √something: (a+b)^3 = 7+√(50). that means a^3+3ab^2=7 and b^3 +3ba^2 = 5√2 a=1 and b=2√2 and the answer of the question is 2.

13:49 Click "Use the real-valued root instead"

You can bypass long division since we know that a=1 in the quadratic and the sum of the x^2 coefficients = 0; -2+b=0 ==> b=2. You then have x^2+2x+c, and we know that c = -14/-2=7. So you have x^2+2x+7 with two complex roots.

WolframAlpha returns the principal root by default, which is the root with the largest real component NOT the purely-real root. There’s a link to get the pure real root near the top of the screen there.

Exactly right. We all know that sqr50=5sqr2 which means we can have 3sqr2+2sqr2. 2sqr2 = (sqr2)^3. In 3sqr2 we have 3 ×sqr2 ×1. And we know that (a+b)^3 = a^3 +b^3 + 3 a^2 ×b + 3 b^2 ×a. In our case we have a^3 which is (sqr2)^3. 3sqr2 ×1 should be 3b^2 ×a becoz if a=sqr2 , 3 a^2 ×b should have been whole number not radical. So b=1 and in 7 we have 1 + 6 which is b^3 + 3a^2 ×b = 1+ 3×(sqr2)^2 × 1. Finally we got ((1+sqr2)^3)^1/3 + ((1-sqr2)^3)^1/3 = 1+1+sqr2 -sqr2=2.

wait, *ARE YOU SECRETLY AN OOD?* for those who don't watch doctor who, a ood is an alien that has a orb in its hand.

It can even be solved by keeping (7+√50)^1÷3 =X+√Y then cube both the sides then make the rational part of lhs=rhs and same with irrational one.

If you wonder how someone came up with that problem, consider this: (1+√2)³ = 1 + 3√2 + 3*2 + 2√2 = 7 + 5√2 = 7 + √50 (1-√2)³ = 1 - 3√2 + 3*2 - 2√2 = 7 - 5√2 = 7 - √50 And obviously, 1 + √2 + 1 - √2 = 2 For the problem setter, all they have to do is start with something of the form a + √b and a - √b, cube them, simplify, and amaze you. You *just* have to notice, going backward, that 7+√50 can be factored as a nice cube.

I am in grade 7 and a similar problem is found in Maths Olympiad homework thanks ❤

Nice advertising for wolframalpha though lol

Hey, I don't really know what is going wrong but if you just press "the real-value root", it will give you the correct answer, you also can do that if you input "cube root of (7+root 50) +cube root of (7-root 50)"

Using formula a^3+b^3=(a+b)(a^2-ab+b^2), you'll directly come to the result, 14/(u^2+3)=u =>u^3+3u-14=0 Where u=(7+(50)^(1/2))^(1/3)+(7-(50)^(1/2))^(1/3)

X=2 is the first zero... Other two zeroes are still possible..... Answers may vary

"If this does not blow your mind, there is something wrong with you" -Salman Khan I think this quote applies here.

I think wolfram alpha is correct because when we raise sth. to the third power, we create extraneous root. In the expression,7 is less than sqrt(50), so the cube root of that should be a+bi.

Mathematica solves this problem instantly: RootReduce[Surd[(7 + Sqrt[50]), 3] + Surd[(7 - Sqrt[50]), 3]]. But it also instantly solves the real-world problem where, for instance, one of the 50's is changed to 50.01. The hand method described here has no hope in solving that! These competition problems are so exquisitely crafted to require some trick, generally very narrow in scope. If you want to address that immensely larger set of problems you'll face in a career in applied math, spend your time instead learning computer algebra.

u need to click on "the real-valued root" on top bro

One of my favourite KZfaq channels!

I decided to pause when you asked, and make an attempt; it was _extremely_ satisfying to work through. Thanks.

Wolfram Alpha says "using the principal root". Instead, click on "using the real-valued root", and you will get 2. Also, if instead of raising the two terms to the 1/3 power, take cuberoot() of them - Wolfram Alpha does have this function, and using that gives 2. Also I note something interesting. As soon as blackpenredpen typed in (7+sqrt(50))^(1/3), the red number below showed 2.414213... That's sqrt(2)+1( !) That may explain this phenomenon of having a complicated expression like (7+sqrt(50))^(1/3)+(7-sqrt(50))^(1/3) turn into just a 2. It's because 7+sqrt(50); i.e., 7+5sqrt(2), is the cube of sqrt(2)+1.

blackpenredpen: Writhes a bunch of cool math and receives a beautiful answer Wolframalpha: That's dope, but how about clicking "the real-valued root instead"

How many comments state this? "Assuming the principal root. Use the real valued root instead." www.wolframalpha.com/input/?i=(7%2Bsqrt(50))%5E(1%2F3)%2B(7-sqrt(50))%5E(1%2F3)&rawformassumption=%22%5E%22+-%3E+%22Real%22

Wolfram Alpha is using the principle branch cut but if you use the command cuberoot or sure(x,n) instead of raising the power to 1/3 you will get your required rational value as x³=c has 3 solutions.

Great vid! Keep up the good work 👍🏻 btw i think your mistake is at 7:10 which was taking (-1) out of root. As you showed us alpha solved this equation in imaginary number because of the inner of the root was negative 😊

If you still don't believe it's 2 Here 7+sqrt(50) = (1 + sqrt(2))³ 7 - sqrt(50) = (1-sqrt(2))³ so, it would be = (1+sqrt(2)) + (1-sqrt(2)) = 1 + 1 + sqrt(2) - sqrt(2) = 2

But dude as it is a cubic equation so it may have 3 factors. So the result of this may be also equal to {-2 ± 2i√6}/2

I can even understand your teaching without audio awesome sir

I recall that somebody suggested a neat way of solving this type of problem: If a + b + c = 0, then a^3 + b^3 + c^3 = 3 abc. (That can be shown by expanding (a + b + c )^3 . ) Then a and b represent the two terms on the left and c represents (-X). Thence : 14 - X^3 = - 3Xab = 3X, so X^3 + 3X -14 =( X - 2)(X^2 + 2X + 7) = 0.

Here in Brazil we call this MATH MAGICS !!!!

I enjoyed this so much! I highly recommend taking a look at photomath's way of solving the problem. In a way, I see it at a true Order of Operations problem as it is solved without the use of variables but still used formulas. It's just an interesting take. I'll link it to you if you are interested.

I don’t have an easy homework as that, in my brain I was thinking that it only can have a complex solution

Using A³ + 3AB(A + B) + B³ for (A + B)³ makes what you did here a little more compact. Written as cbrt(5√2 + 7) - cbrt(5√2 - 7) I saw a trick/method (another KZfaq channel) to expand (a√2 ± b)³ = 5√2 ± 7 and finesse a = b = 1.