problem ( a + b )⁵ = a⁵ + b⁵ Note that ( a + b )⁵ expanded has additional terms besides a⁵ and b⁵. The sum of these must be 0 for our equation to hold. We obtain 5 a⁴ b + 10 a³ b² + 10 a² b³ + 5 a b⁴ = 0 5 a b ( a³ + b³ ) + 10 a² b² ( a + b ) = 0 5 a b ( a + b )(a²-a b + b²)+ 10 a² b² ( a + b ) = 0 Factor. 5 a b (a + b) (a² + ab + b²) = 0 a b (a + b) (a² + ab + b²) = 0 Use the 0 product property. a = 0 b = 0 a = -b b = a e^(-i 2π/3) b = a e^( i 2π/3) answer (a,b) = { (0,0), (0,b), (a, 0), (a,-a), (a, a e^(-i 2π/3) ), (a, a e^(i 2π/3) ) }
@@alexandermorozov2248 и правда. Тогда заменой, но всё равно уравнение конское получается: 5x⁴+10x³+10x²+5x=0. Особенно x³ - самая 3,14 3 до противная переменная.😕🤧
For the complex solutions a/b is -1/2 + i*(3/4)^1/2 or -1/2 - i*(3/4)^1/2, which can also be expressed as e^(i*pi/3) or e^(-i*pi/3). The difference between their arguments is a constant, and they have the same absolute value.
Complex solutions, very interesting. I feel like I’m learning so many fun algebra tricks to factor stuff.
problem
( a + b )⁵ = a⁵ + b⁵
Note that ( a + b )⁵ expanded has additional terms besides a⁵ and b⁵. The sum of these must be 0 for our equation to hold.
We obtain
5 a⁴ b + 10 a³ b² + 10 a² b³ + 5 a b⁴ = 0
5 a b ( a³ + b³ ) + 10 a² b² ( a + b ) = 0
5 a b ( a + b )(a²-a b + b²)+ 10 a² b² ( a + b ) = 0
Factor.
5 a b (a + b) (a² + ab + b²) = 0
a b (a + b) (a² + ab + b²) = 0
Use the 0 product property.
a = 0
b = 0
a = -b
b = a e^(-i 2π/3)
b = a e^( i 2π/3)
answer
(a,b) = { (0,0),
(0,b),
(a, 0),
(a,-a),
(a, a e^(-i 2π/3) ),
(a, a e^(i 2π/3) ) }
Your final answer misses solutions of the form (a, b) = (0, c) and (a,b) = (c, 0) , where c is nonzero.
@ have added them. Yes, (0,0) doesn't cover (a,0) and (0, b) where either a = 0 or b = 0 but not both. Thanks.🙏
@@Don-Ensley No problem! :-)
Взять формулы для пятых степеней, составить из них систему, вычесть... 5a⁴b+10a³b²+10a²b³+5ab⁴=0. Тяжёлое, правда, уравненьице выходит...
И поделить полученное уравнение на b^4, чтобы прийти к уравнению от одной переменной a/b.
@@alexandermorozov2248 а это мысль!
@@alexandermorozov2248 правда, из первого слагаемого, например, снова абракадабра получится, a⁴/b³...🙄
Делить на b^5 (уточнение). Тогда повсюду будут соотношения a/b в разной степени. Поторопился написать, сорри.
@@alexandermorozov2248 и правда. Тогда заменой, но всё равно уравнение конское получается: 5x⁴+10x³+10x²+5x=0. Особенно x³ - самая 3,14 3 до противная переменная.😕🤧
This equation, as stated, has infinitely many solutions. Because if a=0 then b can be any number. The same is true for b=0.
For the complex solutions a/b is -1/2 + i*(3/4)^1/2 or -1/2 - i*(3/4)^1/2, which can also be expressed as e^(i*pi/3) or e^(-i*pi/3). The difference between their arguments is a constant, and they have the same absolute value.
It's e^(±i*2π/3) (or alternatively written as -e^(∓i*π/3) ), instead of e^(±i*π/3) (which have _positive_ real parts).