Your second line is incorrect.

Very nice.

interesting 👏👏

👌👌👌

Excellent

Thank you for the video. I don't understand why you supposed that a,b are both even when they are clearly not as (a^b-b^a) is odd.

@2:00 why are x and y necessarily integers?

Isn't a = 18, b = 1 a (very obvious) solution too?

Great❤

Beautiful

lol

thanks bor

15^30

The answer is 15^30. You can't cancel exponentiation the same way you cancel divisors.

My approach is like this 30^60/60^30 , here we can note that power of 30 is n×2 the power of 60 , so on that case 30^2/60^1 = 900/60 = 15 So if it was 30^60 / 60^30 then it will be eqaul to 15^30 as 30^2 / 60^1 = 15 , also 30^4 / 60^2 = 15^2 hence 30^60/60^30 = 15^30

or you could 30th root the whole thing, leaving you with 30^2/60, then simplify 90/60 to 3/2

This is inaccurate: There are 4 solutions to this exercise and not only two...: Two of them are real: 1) X+Y = 9 and 2) X+Y=-9 and two of them are complex (imaginary): 3) X+Y = 3i 4) X+Y= - 3i i = is the square root of minus 1

(x+y)(x-y) = 27. This suggests x+y = 9, x-y =3. So 2x = 12, x = 6, y =3

great sir😀

seriously, math olympiad question? first desicion (without math at all) multiplying x*y = -15 it means first (x or y) is with - second (x or y) with + x-y = 8, means x greater than y, so x with + sign y with - sign multiplying gives us nonpair integer and dividing gives us pair integer it means that x is positive nonpair integer y is negative nonpair integer not so hard to solve x = 5 y = -3 5^4 +(-3)^4 = 706 second desicion x - y = 8 => x=8+y y*(8+y) + 15 = 0 y^2 +8*y +15 = 0 y = -3 x=8+y = 5 5^4 +(-3)^4 = 706 why the hell we are playing with formulas when answer is so obvious?

brilliant

1?

Thx 😎👍

Hi! I don't quite understand... I achieved to isolate the x's with a polynomial which equals to 0 (--> 48x^2-16x+1=0). And i found two roots : 1/4 and 1/12. Of course 1/12 doesn't work, but I don't understand why. Could you explain this to me please ? Thanks

the solution was damn good

missed 2nd root, if t=-5 x=1/12

If you cannot notice such elegant polynomial transformations and factorizations. Let f(x) = x⁹+x⁶-36 => f'(x) = x⁵(9x³+6) => f'(x) < 0 <=> x is in interval (-⅔^⅓, 0) => for all x < 0 f(x) <= f(-⅔^⅓) < 0 (since the function is increasing until -⅔^⅓ and then decreasing afterward until 0.) => there are no solutions for x < 0. Notice that x = 0 isnt a solution. If x > 0 then f'(x) > 0 which implies there can be at most one solution (if the function is always increasing it can intersect x axis at most once) By playing around you can see that 3³+3² = 36. (i first tried out 1, and then 2, and then 3...) Hence x³ = 3 <=> x= 3^⅓ is the only real solution.

4:00, I would preffer "no real solution"

When the discriminant is less than 0 it doesn’t mean it’s impossible it just means that there’s no real solutions. You still have imaginary solutions

It's amazing how easy it is to solve, but at the same time how difficult it is

Dahm

Don’t the exponent add up to 6? Since you multiplied everything

Helpful vid, thanks Aram!

interesting 👏👏👏

please subscribe if you liked it🤍it help me a lot and make me happy😇🤍

Thats actually brilliant

❤

Should be able to do this without pen and paper, let alone a calculator.

Bro I got the answer in like 2-3 steps

Close your brackets. 😬

Straightforward👍

Nice👌

👍🏻

Nice

Nice

It was very useful❤