Minimum requirement: tell us the underlying set (when presenting the problem) where x and y are from. Integers? Non-negative integers? Positive integers? Reals? Rationals? Complex numbers??? Each of these is a different problem that suggests a different set of tools to use.

Did the question even indicate “integers”? If not there’s infinite solves….

@somacruzin4474

Жыл бұрын

💀

@pinoes31416

11 ай бұрын

it did not. As you state, infinite solutions as you chose (n^3, 91+n^3) numbers...

If I'm ever given a problem presented like this on a exam, I answer that there are infinite solutions. In fact for every number x, there is a corresponding y that satisfies it. You can even present it as a function: y = cuberoot(x^3 - 91). And I expect full marks because that's 100% correct and it's not my fault they forgot to specify the numbers have to be integers.

The more complete solution is to let u=x-y, then divide both sides by u=x-y. A quadratic equation on the left side occurs: x^2+xy+y^2=91/u. Substitute x=y+1 for x on the left side to generate a quadratic equation involving y and parameter u. Solve quadratic equation for y as a function of parameter u. The value of x is y+u. This generates a family of solutions for a parameter u. If you want integer solutions, select an integer parameter u that divides 91. You can also pick u to generate rational number solutions. It depends of the field you want, integers, rational numbers, complex numbers etc. This is the advantage of a parametrized solution.

Aren’t there an infinite number of solutions unless you specifically state that the solutions have to be whole numbers? I guess that’s implied? Otherwise u can just plug in any value for y and x=cbrt(91+y^3).

When you get to (y+1)y = 30 you can easily easily find two numbers that differ by 1 and multiply out to 30 without getting a quadratic equation. 5 & 6 differ by 1 & multiply out to 30 so y=5 and -6 & -5 differ by 1 and multiply out to 30 so y=-6 is also an answer.

The answers found (6,5) (-5,-6) and (4,-3) (3,-4) are the difference and sum of cubes respectively. Initially I visualized the difference of 2 cubes before watching. The difference of the cube sides, delta, gives a volume difference between the cubes of 91. Taking the smaller cube side, a, (absolute), the size of the volume difference is 3 x a^2 x delta + 3 x a x delta^2 + delta^3 = 91. In order to do the problem mentally I assumed delta was a whole number. Considering a^2 x delta (I didn't consider the sum of cubes. ) Once the integer cubes cases are found, the equation for all real solutions has example test cases.

At a quick glance I start by calculating cubes and their differences . 4 cubed = 64, 5 cubed = 125 and 6 cubed = 216 then 216 - 125 = 91 hence 6^(3) - 5^(3) = 91 and x= 6 and y = 5.

did it in 20 seconds by hit and trial x=6 y=5

@mmm2233

Жыл бұрын

lol

@anka2k6

Жыл бұрын

same lol

@higher_mathematics

Жыл бұрын

Great, Sir!

@goutamboppana961

Жыл бұрын

and x=4 y=-3

@alexb6695

8 ай бұрын

Exactly. 1st I tried 7 n 6 n straight away realised a bit to much, so next step was 6 n 5. Can't believe he used full A4 to explain it😂

Not sure what level this one is, but my guess is pretty much similar to the Cambridge Local Examination Syndicate GCE O' level that I passed aged 14!

@Mathemagical55

Ай бұрын

O-Level papers were more difficult than A-levels today

Any solution in (x,cbrt(x^3-91)

X=6. Y=5 6^3-5^3=91

4 and -3 or 6 and 5 both work

Would have been easier to just pick a number and plug into the equation until you get the right answer

x= -0.5 and y=-4.5, this is also a possibility?

Solution x plus y equals A

X=6,y=5

6^3-5^3=91

4,-3

Again, I did this very directly. You can often do this for integers when the numbers are small. If you make a little table of cubes you get 0 0 1 1 2 8 3 27 4 64 5 125 6 216 7 343 8 512 Now make table of 91 - z^3 (y negative) and 91+z^3 (y positive) 0 91 91 1 90 92 2 83 99 3 64 118 (64 is a cube, see handy table above) 4 27 155 (27 is a cube) 5 -34 216 (216 is a cube) 6 -125 307 (-125 is a cube) 7 -252 434 This is already enough You can read off the answers by spotting the cubes, 64, 27, -125, and 216 After that the gaps are too big to fit more solutions. So the answers come at y=-3, y=-4, y=-6, y=5 the corresponding roots x=4, x=3, x=-5, x=6

@lucapolidori8817

6 ай бұрын

Interesting enoguh, I think: put in another column the differences between x^3 and (x-1)^3. You'll find 1 , 7, 19, 37, 61,91.... this serie increments by 6, 12, 18, 24, 30...If you want to find the difference between two cubes, you just have to start from the multiples of 6.

X=6 y=5

X7*y13

I did not get the part where u said xy would always be positive and x-y could be -ve or +ve also why did you reject the cases with the negative numbers?

@tlemarvel9505

Жыл бұрын

In the cases where both x and y were negative, xy would ALWAYS be positive. The square of a negative number will ALWAYS be positive as well. Therefore eliminating the possibility of those options.

@VenturaNotes

Жыл бұрын

(x^2 + xy + y^2) will always be positive given that x and y both don't equal to 0. This is because the sum of squares (x^2+y^2) will always be positive and greater than the product of their bases (xy). (x-y) can be negative or positive because x and y could both be positive number or negative numbers in this case. The cases with the negative numbers were rejected because it would require (x^2+ xy +y^2) to be negative which is impossible as explained above.

@higher_mathematics

Жыл бұрын

Great approach, Sir!

@tallbell

8 ай бұрын

Sorry, I still havn't got this point. If x=-1 and y=-91 then x*y = (-1)*(-91)=91 then all elements are positive and sum also positive🤔

@johnbrennan3372

7 күн бұрын

X^2 +XY+Y^2= ( X-Y)^2 +3XY.Therefore X^2 +XY+ Y^2 is greater than X-Y which means while 1,91 works, 91,1 doesn’t work etc.

Thank you for the problem

@higher_mathematics

Жыл бұрын

Thank You for watching, Sir! Have a nice day!

jee laughing in the corner

@somacruzin4474

Жыл бұрын

Ruk ja bhai, results ke baad crying in the corner hoga 😢

@vishulkhinchi2007

Жыл бұрын

@@somacruzin4474 mera to nhi hoga Bhai

Z always ask X why ! This is nonsense 🧐

Beautiful sulation

x is 6. y is 5

@higher_mathematics

Жыл бұрын

Not only this roots. 3 more pairs. Thank you for watching

@somacruzin4474

Жыл бұрын

​@@higher_mathematics (-5,-6) (6,5) (4,-3) (3,-4) I used the substitution method, so x - y = 1 => x = y +1, and I put the value of x in x² + xy + y² = 91

@somacruzin4474

Жыл бұрын

​@@higher_mathematics however, when I change the values, say, x - y = 91 and x² + xy + y² = 1 I end up with a equation : 3y² + 273y + 8280 = 0 I tried solving it in the method I was taught in 8th grade but it's tooooo big. So just now it hit my head to use the quadratic formula. I'll let you know if the results differ or not

@somacruzin4474

Жыл бұрын

​@@higher_mathematics ok done, so if we reverse the values the solutions will be imaginary not real

@somacruzin4474

Жыл бұрын

Also feedback on video, I have a short attention span, so I just started doing the problem on my own after you drew the first curly bracket

im not supposed to be here, i took law and not whatever is this that i learn in HS. (i know what is this but my brain decided to shutdown). anyway, thanks for refreshing my brain.

@higher_mathematics

Жыл бұрын

Thank You, Sir!

It’s obvious at a glance that x = 6 and y = 5. Try it out, no need to calculate what doesn’t need to. 6^3=216 5^3=125 216-125=91

It took me a while to put it into words, but i got it nearly instantly viewing it from an electric signals perspective. If you look at x^3 and y^3 as plots of functions instead of as variables, and think about in what scenario substracting one function to another function always results in the same constant, then that means both functions are the same, except that one has an added constant. So x(n) = n + 91, and y(n) = n. But wait, what about the ^3? well just give your functions an umbrella, x(n) = cube_root(n+91) and y(n) = cube_root(n). If this is a multiple choice question, just replace n so that its cube root equals one of the given choices of values for y and check if the given value for x matches your x(n)

@somacruzin4474

Жыл бұрын

Teach me this skill 🙏

If I'm ever given a problem presented like this on a exam, I answer that there are infinite solutions. In fact for every number x, there is a corresponding y that satisfies it. You can even present it as a function: y = cuberoot(x^3 - 91). And I expect full marks because that's 100% correct and it's not my fault they forgot to specify the numbers have to be integers.