My method, raise to power 1/6 both sides ---> a^(1/3) * b^(1/2) = 6 ---> b^(1/2) must be divisor of 6 ---> b^(1/2) = 1, 2 , 3, 6 ----> b = 1, 4, 9, 36 substitute and you have corresponding values of a
You have done math proud by including complex solutions in many of these problems. Can you do a series that includes quaternion solutions to many of these problems? I recognize that this particular one is limited to integers.
a^2 * b^3 = 6^6. Now raise both sides to the power of 1/6. cbrt(a) * sqrt(b) = 6. I'll divide both sides by cbrt(a) and then square both sides to get: b = [6 / cbrt(a)]^2. Notice that b has to be an integer, so the RHS is also an integer. Therefore cbrt(a) has to be a divisor of 6. This gives us cbrt(a) = +-1, +-2, +-3, +-6. Therefore a = +-1, +-8, +-27, +-216. And if you put these into the formula for b, we get b = 36, 9, 4, 1. Therefore the solutions are: (a,b) = {(1, 36),(-1,36),(8,9),(-8,9),(27,4),(-27,4),(216,1),(-216,1)}.
I found it simpler just to note that 6^6 = 2^6 * 3^6 OR 1^6 * 6^6. Either way, we have two factors on the left and two factors on the right. So, just set them equal to each other and find all the combinations. E.g., let a^2 = 2^6. Solve for a. Then b^3 = 3^6. Solve for b. Then let a^2 = 3^6. Etc. It's pretty quick, and doesn't even require paper.
@SyberMath Have I listed all the solutions in my comment? No, obviously not. That's why I wrote, "Etc." Does this method yield all the solutions when followed all the way? Yes, I believe so.
a²b³ = 6^6 = (2² 3²)³
a² = (2² 3² / b)³
As only integer solutions are admitted, the right side of the equality must be a perfect square, since the left side is equal to a². Then:
b = 1 ==> a² = (2² 3²)³ ==> a = ±(2 . 3)³ = ± 216
b = 2² ==> a² = (3²)³ ==> a = ±3³ = ± 27
b = 3² ==> a² = (2²)³ ==> a = ±2³ = ± 8
b = 2² 3² ==> a² = 1³ ==> a = ±1³ = ± 1
All solutions: (-216 , 1) ; (216 , 1) ; (-27 , 4) ; (27 , 4) ; (-8 , 9) ; (8 , 9) ; (-1 , 36) ; (1 , 36)
My method, raise to power 1/6 both sides --->
a^(1/3) * b^(1/2) = 6 ---> b^(1/2) must be divisor of 6 ---> b^(1/2) = 1, 2 , 3, 6 ----> b = 1, 4, 9, 36
substitute and you have corresponding values of a
Great and simple way of solution!
You have done math proud by including complex solutions in many of these problems. Can you do a series that includes quaternion solutions to many of these problems?
I recognize that this particular one is limited to integers.
Very nice solutions. Eight in all, neatly found.
a=±8, b=9; a=±27, b=4
Nice! 👍
@@alexandermorozov2248 kittos
B=36, a=1
a^2 * b^3 = 6^6. Now raise both sides to the power of 1/6. cbrt(a) * sqrt(b) = 6. I'll divide both sides by cbrt(a) and then square both sides to get: b = [6 / cbrt(a)]^2. Notice that b has to be an integer, so the RHS is also an integer. Therefore cbrt(a) has to be a divisor of 6. This gives us cbrt(a) = +-1, +-2, +-3, +-6. Therefore a = +-1, +-8, +-27, +-216. And if you put these into the formula for b, we get b = 36, 9, 4, 1. Therefore the solutions are: (a,b) = {(1, 36),(-1,36),(8,9),(-8,9),(27,4),(-27,4),(216,1),(-216,1)}.
I found it simpler just to note that 6^6 = 2^6 * 3^6 OR 1^6 * 6^6. Either way, we have two factors on the left and two factors on the right. So, just set them equal to each other and find all the combinations. E.g., let a^2 = 2^6. Solve for a. Then b^3 = 3^6. Solve for b. Then let a^2 = 3^6. Etc. It's pretty quick, and doesn't even require paper.
Are those all the solutions?
@SyberMath Have I listed all the solutions in my comment? No, obviously not. That's why I wrote, "Etc." Does this method yield all the solutions when followed all the way? Yes, I believe so.
(2*3)^2=6^2 per la regola dei prodotti notevoli di moltiplicazione
a = 8, b = 9 / a = 27, b = 4
a = 216, b = 1 / a = 1, b = 36
If two equations having two variables and same same coefficients of the variables then the lines are same.
Genial, m' encantan les Equacion Diofàntiques 🐾🎸
8^2х9^3
a=8, b=9
Right, the easy one is by inspection.
There are some more solutions.
I prime factored and noticed that b^3 can only be 1,2^6,3^6 and 6^6
I do not understand how to do this one.
Як завжди 🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
Дякую.
a=8,&b=9, don't ask why
Why did he define a² = 2^x×3^y?
👍
a=1, b=36 :)
or
a=216, b=1
Got 'em all!