$y-1$^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 $thishasnorealsolution$ or y = ±1

Therefore, the solutions to the equation are:

y = 1 or y = -1