x^3 - y^3 = 91

Рет қаралды 10,058

Күн бұрын

Пікірлер: 38

@kausarali3292 Ай бұрын

Wow wonderful solution! Your every question is awesome. Your way of teaching is mesmerizing! And most importantly your cute smile is fabulous and cute! Lot's of love from Pakistan

@yaseenshahin1404 Ай бұрын

The flexing part sent me 😂

@KeshavThakur-ej4zd Ай бұрын

Can you make a video on Riemann's hypothesis .....

@srisaishravan5512 Ай бұрын

Ayy, my favourite youtuber uploas again. I wrote a test yesterday and we had this same question. Well, im still gonna watch it. Keep going sir!

@pojuantsalo3475 Ай бұрын

My solution: Since both x and y are positive integers, y ≥ 1 and y³ ≥ 1. This means that x³ ≥ 1 + 91 = 92. Since 4³ = 64 and 5³ = 125, we have x ≥ 5. Now assume x = 5 and solve for y: y³ = 5³ - 91 = 125 - 91 = 34 which doesn't give integer value for y. x ≠ 5. Next assume x = 6: y³ = 6³ - 91 = 216 - 91 = 125 = 5³. We have a solution x = 6, y = 5. Next assume x = 7: y³ = 7³ - 91 = 343 - 91 = 252 ≈ 6.31636³ which doesn't give integer value for y. x ≠ 7. Also, the difference of cubes of successive integers has grown too large: 7³ - 6³ = 127 > 91. This means there can't be any other solutions for x > 7. x = 6, y =5 is the only solution.

@vamsikirilenko Ай бұрын

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@Brixman 16 күн бұрын

We love you, Math Bob Ross.

@KookyPiranha Ай бұрын

You can omit a lot of unnecessary computation as shown at 3:09: $x^3-y^3=(x-y)(x^2+xy+y^2)$ Note that for $x,y\in\mathbb{N}, x-y=\sqrt{x^2+y^2-2xy}

@cret859 Ай бұрын

Nice video. I like the sincerity with which the explanations are given and especially how doubts are expressed as the solution progresses. If negative integers are allowed, then I find that the solutions of x³-y³=91 are the following (x,y) pairs (-5,-6) (3,-4) (4,-3) (6,5). The only fully positive pair is indeed (6,5).

@christophedesaubliaux5223 Ай бұрын

Your writing is gorgeous too!

@oida10000 Ай бұрын

x^3-y^3=91 | +y^3 x^3=y^3+91 Note that x^30 as this is the difference and this always the case the determinate of the quadtric is negative. So x^3=y^3+9191 --> x^2+x>30, dividing by 2 we see that (x^2+x)/2>15 and as this is known as the xth triangular number and for such small values should be known: x>5.

@katlegotshabalala6250 Ай бұрын

Great problem...smart restriction because if there wasn't, all the x y values are valid for this problem Great!!

@dan-florinchereches4892 Ай бұрын

Wow i like how the problem looks deceptively simple. Obviously x>y is a condition to be fulfilled Rewriting (x-y)(1/2x^2+1/2y^2+1/2(x+y)^2)=91 This means we can only accept positive values for each factor. Now we can simplify as (X-y)(X^2+y^2+(x+y)^2)=182 Since x and y are integers the factor containing squares will always be larger than x-y Compare x-y with x^2+y^2+(x+y)^2 Multiply by x+y and the left is less than the right unless y=-x in which case they are equal but the original integer is not a multiple of 2 so we don't have to consider this possibility Factoring 182=2*91=2*7*23 we can have the following possibilities X-y=1 X^2+y^2+(x+y)^2=182 So (1+y)^2+y^2+(2y+1)^2=182 6y^2+6y+2=182 6y^2+6y-180=0 Y^2+y-30=0 (Y+6)(Y-5)=0 => (x,y)€{(-5,-6),(6,5)} Or X-y=2 X^2+y^2+(x+y)^2=182 (Y+2)^2+y^2+(2y+2)^2=182 6y^2+12y+8=182 6y^2+12y-174=0 Y^2+2y-29=0 no integer solutions X-y=7 (Y+7)^2+y^2+(2y+7)^2=182 6y^2+42y+98=182 6y^2+42y-84=0 Y^2+7y-14=0 no integer solutions X-y=14 (Y+14)^2+y^2+(2y+14)^2=182 6y^2+3*2*14y+196+196=182 6y^2+6*14y^2+196+14=0 Y^2+14y+35=0 no integer sol

@DaiNguyen-dj1ci Ай бұрын

Awesome video! 😊

@nicolasb11 Ай бұрын

Well done I love the way you're doing fun buy teaching us maths 😊

@yurenchu Ай бұрын

(x-1)³ = x³ - 3x² + 3x - 1 = = x³ - 3*(x² - x + 1/3) = x³ - 3*(x² - x + 1/4 + 1/12) = x³ - 3*( (x - 1/2)² + 1/12 ) x³ - (x-1)³ = 3*( (x - 1/2)² + 1/12 ) Note that righthandside is always positive, for any real value of x . So if |x - 1/2| > 5.5 , then 3*( (x - 1/2)² + 1/12 ) > 3*( (5.5)² + 1/12 ) = 3*(30.25 + 1/12) = 91 which means the difference between consecutive cubes (x-1)³ and x³ will be greater than 91 . Therefore, |x - 1/2| ≤ 5.5 ==> -5 ≤ x ≤ 6 x³ - y³ = 91 x³ = 91 + y³ y³ = x³ - 91 x x³ (x³ - 91) -5 -125 -216 = (-6)³ -4 -64 -155 , is not a cube -3 -27 -118 , is not a cube -2 -8 -99 , is not a cube -1 -1 -92 , is not a cube 0 0 -91 , is not a cube 1 1 -90 , is not a cube 2 8 -83 , is not a cube 3 27 -64 = (-4)³ 4 64 -27 = (-3)³ 5 125 34 , is not a cube 6 216 125 = (5)³ Integer solutions: (x, y) = (-5, -6) , (3, -4), (4, -3) or (6, 5)

@itsphoenixingtime Ай бұрын

Very nice but they asked for x y both positive Extra credits for finding all integer pairs though.

@yurenchu Ай бұрын

@itsphoenixingtime Somehow I hadn't noticed the word "positive" in the thumbnail. It's only after I posted this comment and played the video that I noticed that the question on the blackboard says "positive integers". If I had known, I could have limited my "table" search to integer values of x ≤ 6 such that x^3 ≥ 91 ; in other words, x = 5 or 6 .

@nothingbutmathproofs7150 Ай бұрын

A cleaner version would be when you got to (y+1)*3y = 90 => (y+1)y=30. A number (y) times one more than that number (y+1) = 30 => y=5 and y+1 = 6 (so y=5).(or y=-6) You don't need to both with a quadratic equation

@refael8525 Ай бұрын

Started watching this to as a background video now I’m in too and started watching all the videos I can get my hands on

@flexeos Ай бұрын

at first, to be comprehensive, you should also consider -1,-91 and -7,-13.and then eliminate them because if x,y>0 then x2+xy+y2>0

@samuelgomes4288 26 күн бұрын

Actually x²+xy+y² is non negative for any reals x, y: (x/y)²+(x/y)+1 > 0 (you can verify) Multiply by y²: x²+xy+y² >= 0 also including (0, 0)

@anestismoutafidis4575 Ай бұрын

6^3 -5^3=216-125=91; (216^1/3)-(125^1/3)=91; (3!)^3-(3!-1!)=91; 3^3-(-4)^3=27+64=91; 3log(10^31)- 3log8(4)=93 -2=91; 3ln(100×10^30)- 3ln(81×20^13)=221 -130=91

@itsphoenixingtime Ай бұрын

So I tried to solve it by factoring. The issue is that: 1. I found out that there is no point in trying to take double negative factors, i.e where x-y = negative and x^2 + xy + y^2 = negative. Since x and y are positive, x^2, xy and y^2 are positive. So as a result you cannot have the second factor as a negative. Moreover the only factor pairs possible are 1,91 and 7,13. (Technically there are the reversed cases, but you could always prove that x^2 + xy + y^2 is always larger than x-y, as it is squared + 3xy, but for the sake of completeness i checked the cases) 1,91 produces a solution x = 6, y = 5, and another one that has x has negative. 7,13 produces 2 solutions with positive x and negative y. So they are out. Only solution is 6,5 Though to be honest it would be interesting to see for ALL integers and not just positive. But that means taking up to 8 different cases.

@maths01n 25 күн бұрын

Great ❤

@rob876 Ай бұрын

x^3 - y^3 = (x - y)(x^2 + xy + y^2) = 91 for one solution, x - y = 1, x^2 + xy + y^2 = 91 x = y + 1, x^2 + x(x - 1) + (x - 1)^2 = 91 3x^2 - 3x - 90 = 0 x^2 - x - 30 = 0 (x - 6)(x + 5) = 0 x = 6, y = 5 for the other, x - y = 7, x^2 + xy + y^2 = 13 x^2 + x(x - 7) + (x - 7)^2 = 13 3x^2 - 21x + 36 = 0 x^2 - 7x + 12 = 0 (x - 3)(x - 4) = 0 y will be < 0 in both cases

@Uranyus36 Ай бұрын

great video! I regret not remembering the last question of my analysis exam back in university which I really want to know the solution. It's proving something given a relation about composition of a function. If I wrote down the problem I could've given you to solve for me xd

@paulortega5317 Ай бұрын

Or (x - y)( (x - y)² + 3xy) = 91 (x - y) < ∛91 (x - y) < 5 That leaves (x - y) = 1 from the set of possible factors (1, 7, 13, 91) (1 + 3xy) = 91 xy = 30 and only solution is (6,5)

@walterwen2975 Ай бұрын

x³ - y³ = 91; Find all positive integers of x, y x³ > y³ > 91, x, y ϵ ℤ: x = 4, y³ = x³ - 91 = 4³ - 91 = - 27; x > 4 x³ - y³ = 91 = (7)(13), 7, 13 are prime numbers; 7 > x > 4 x³ - y³ = (7)(13) = (6 + 1)(36 - 23) = 216 + 6(6 - 23) - 23 = 216 - 125 = 6³ - 5³ x³ - y³ = 6³ - 5³; x = 6, y = 5; No more positive integer root for x, y Answer check: x³ - y³ = 91; Confirmed as shown Final answer: x = 6, y = 5

@Maths786 Ай бұрын

Sir please do a limit question which was came in JEE Advanced 2014 shift-1 question number 57 it's a question of a limit lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢