Question

X+y+z=1 x2+y2+z2=4 x3+y3+z3=6 x4+y4+z4=?

Answer 1

x + y + z = 1  (1)

x^2 + y^2 + z^2 = 4 (2)

x^3 + y^3 + z^3 = 6  (3)

x^4 + y^4 + z^4 = ?

не совсем все просто

надо очень хорошо знать квадраты и кубы многочленов

Итак

(x + y)^2 = x^2 + 2xy + y^2

(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3

(x + y + z)^2 = (x + y)^2 + 2(x + y)z + z^2 = x^2 + y^2 + z^2 + 2(xy + yz + xz) (4) аналогично и квадраты

(x^2 + y^2 + z^2)^2 = (x^2 + y^2)^2 + 2(x^2 + y^2)z^2 + z^4 = x^4 + y^4 + z^4 + 2(x^2y^2 + y^2z^2 + x^2z^2) (5)

рассмотрим 4

x^2 + y^2 + z^2 + 2(xy + yz + xz)  = 1

2(xy + yz + xz)  = 1 - 4

xy + yz + xz  =  - 3/2

(x + y + z)^3 = (x + y)^3 + 3(x + y)^2z + 3(x + y)z^2 + z^3 = x^3 + 3x^2y + 3xy^2 + y^3 + 3x^2z + 6xyz + 3y^2z + 3xz^2 + 3yz^2 + z^3

6xyz = (x + y + z)^3 - 3(x + y + z)(x^2 + y^2 + z^2) + 2(x^3 + y^3+z^3)

6xyz = 1 - 3*1*4 + 2*6 = 1 - 12 + 12 = 1

xyz = 1/6

и наконец

(x^2 + y^2 + z^2)^2 = (x^2 + y^2)^2 + 2(x^2 + y^2)z^2 + z^4 = x^4 + y^4 + z^4 + 2(x^2y^2 + y^2z^2 + x^2z^2) (5)

x^4 + y^4 + z^4  = (x^2 + y^2 + z^2)^2 - 2(x^2y^2 + y^2z^2 + x^2z^2) = (x^2 + y^2 + z^2)^2 - 2((xy + yz + xz)^2 - 2xyz(x+ y + z))  

получили окончательную формулу

x^4 + y^4 + z^4  =  (x^2 + y^2 + z^2)^2 - 2((xy + yz + xz)^2 - 2xyz(x+ y + z))  

x^4 + y^4 + z^4  =  4^2 - 2((-3/2)^2 - 2*1/6*1)  = 16 - 2(9/4 - 1/3) = 16 - 2(27 - 4)/12 = 16 - 23/6 = 16 - 3 5/6 = 12 1/6 = 73/6

ответ 73/6

проверьте - вроде правильно

Answer 2

To find the value of x^4 + y^4 + z^4, we can use the following approach:

First, let's solve the given system of equations:

Equation 1: x + y + z = 1 Equation 2: x^2 + y^2 + z^2 = 4 Equation 3: x^3 + y^3 + z^3 = 6

Now, let's square Equation 1 and simplify it:

(x + y + z)^2 = 1^2 x^2 + y^2 + z^2 + 2(xy + xz + yz) = 1

Using Equation 2, we can substitute the value of x^2 + y^2 + z^2:

4 + 2(xy + xz + yz) = 1 2(xy + xz + yz) = 1 - 4 2(xy + xz + yz) = -3 xy + xz + yz = -3/2

Next, let's cube Equation 1 and simplify it:

(x + y + z)^3 = 1^3 x^3 + y^3 + z^3 + 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 6xyz = 1

Using Equations 2 and 3, we can substitute the values:

6 + 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 6xyz = 1 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 6xyz = 1 - 6 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 6xyz = -5

Using the values of xy + xz + yz from earlier, we can substitute:

3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 6xyz = -5 3(x^2y + xy^2 - (3/2)(x + y + z)z) + 6xyz = -5 3(x^2y + xy^2 - (3/2)(1)z) + 6xyz = -5 3(x^2y + xy^2 - (3/2)z) + 6xyz = -5 3(x^2y + xy^2 - (3/2)z) + 6xyz = -5

We now have a system of equations that can be solved for the values of x, y, and z.

From here, we can calculate x^4 + y^4 + z^4 using the values of x, y, and z obtained from the system of equations. However, since the calculations involve higher powers, they may not result in simple integer values. To find the exact value of x^4 + y^4 + z^4, you would need to solve the system of equations and perform the necessary calculations.