please slove for the variable, and make the world smile again?
x^4+4x^2−32=0
t^6=8−t^3
thank you for your greatness
Solve for x:
x^4+4 x^2-32 = 0
Substitute y = x^2:
y^2+4 y-32 = 0
The left hand side factors into a product with two terms:
(y-4) (y+8) = 0
Split into two equations:
y-4 = 0 or y+8 = 0
Add 4 to both sides:
y = 4 or y+8 = 0
Substitute back for y = x^2:
x^2 = 4 or y+8 = 0
Take the square root of both sides:
x = 2 or x = -2 or y+8 = 0
Subtract 8 from both sides:
x = 2 or x = -2 or y = -8
Substitute back for y = x^2:
x = 2 or x = -2 or x^2 = -8
Take the square root of both sides:
Answer: | x = 2 or x = -2 or x = (2 i) sqrt(2) or x = (-2 i) sqrt(2)
Solve for t over the real numbers:
t^6 = 8-t^3
Subtract 8-t^3 from both sides:
t^6+t^3-8 = 0
Substitute x = t^3:
x^2+x-8 = 0
Add 8 to both sides:
x^2+x = 8
Add 1/4 to both sides:
x^2+x+1/4 = 33/4
Write the left hand side as a square:
(x+1/2)^2 = 33/4
Take the square root of both sides:
x+1/2 = sqrt(33)/2 or x+1/2 = -sqrt(33)/2
Subtract 1/2 from both sides:
x = sqrt(33)/2-1/2 or x+1/2 = -sqrt(33)/2
Substitute back for x = t^3:
t^3 = sqrt(33)/2-1/2 or x+1/2 = -sqrt(33)/2
Take cube roots of both sides:
t = (sqrt(33)/2-1/2)^(1/3) or x+1/2 = -sqrt(33)/2
Subtract 1/2 from both sides:
t = (sqrt(33)/2-1/2)^(1/3) or x = -1/2-sqrt(33)/2
Substitute back for x = t^3:
t = (sqrt(33)/2-1/2)^(1/3) or t^3 = -1/2-sqrt(33)/2
Take cube roots of both sides:
Answer: | t = (sqrt(33)/2-1/2)^(1/3) or t = -(1/2+sqrt(33)/2)^(1/3
