Решите ураанение: (y-1)^4+(y+1)^4=16

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(y-1)^4 + (y+1)^4 = 16

Expand each term using the formula for the fourth power of a binomial:

(y-1)(y-1)(y-1)(y-1) + (y+1)(y+1)(y+1)(y+1) = 16
(y^2 - 2y + 1)(y^2 - 2y + 1) + (y^2 + 2y + 1)(y^2 + 2y + 1) = 16
(y^4 - 4y^3 + 6y^2 - 4y + 1) + (y^4 + 4y^3 + 6y^2 + 4y + 1) = 16

Combine like terms:

2y^4 + 12y^2 + 2 = 16
2y^4 + 12y^2 - 14 = 0

Divide the whole equation by 2 to simplify:

y^4 + 6y^2 - 7 = 0

Let x = y^2:

x^2 + 6x - 7 = 0

Now, solve this quadratic equation for x:

(x + 7)(x - 1) = 0
x = -7 or x = 1

Since x = y^2:

y^2 = -7 or y^2 = 1

Taking the square root on each side:

y = √-7 (this has no real solution) or y = ±1

Therefore, the solutions to the equation are:

y = 1 or y = -1