Question

How do you solve x for (x+5)^3/2 = ( x-1)^3

Answer 1

Answer:

x = 6 (2^(1/3) + 2^(2/3)) + 7 or x = 6 (-2)^(1/3) ((-1)^(1/3) - 2^(1/3)) + 7 or x = 6 (-2)^(1/3) ((-2)^(1/3) - 1) + 7

Step-by-step explanation:

Solve for x:

1/2 (x + 5)^3 = (x - 1)^3

Expand out terms of the right hand side:

1/2 (x + 5)^3 = x^3 - 3 x^2 + 3 x - 1

Subtract x^3 - 3 x^2 + 3 x - 1 from both sides:

1 - 3 x + 3 x^2 - x^3 + 1/2 (x + 5)^3 = 0

Expand out terms of the left hand side:

-x^3/2 + (21 x^2)/2 + (69 x)/2 + 127/2 = 0

Bring -x^3/2 + (21 x^2)/2 + (69 x)/2 + 127/2 together using the common denominator 2:

1/2 (-x^3 + 21 x^2 + 69 x + 127) = 0

Multiply both sides by 2:

-x^3 + 21 x^2 + 69 x + 127 = 0

Multiply both sides by -1:

x^3 - 21 x^2 - 69 x - 127 = 0

Eliminate the quadratic term by substituting y = x - 7:

-127 - 69 (y + 7) - 21 (y + 7)^2 + (y + 7)^3 = 0

Expand out terms of the left hand side:

y^3 - 216 y - 1296 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:

-1296 - 216 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

z^6 + z^4 (3 λ - 216) - 1296 z^3 + z^2 (3 λ^2 - 216 λ) + λ^3 = 0

Substitute λ = 72 and then u = z^3, yielding a quadratic equation in the variable u:

u^2 - 1296 u + 373248 = 0

Find the positive solution to the quadratic equation:

u = 864

Substitute back for u = z^3:

z^3 = 864

Taking cube roots gives 6 2^(2/3) times the third roots of unity:

z = 6 2^(2/3) or z = -6 (-1)^(1/3) 2^(2/3) or z = 6 (-2)^(2/3)

Substitute each value of z into y = z + 72/z:

y = 6 2^(1/3) + 6 2^(2/3) or y = 6 (-1)^(2/3) 2^(1/3) - 6 (-1)^(1/3) 2^(2/3) or y = 6 (-2)^(2/3) - 6 (-2)^(1/3)

Bring each solution to a common denominator and simplify:

y = 6 (2^(1/3) + 2^(2/3)) or y = 6 (-2)^(1/3) ((-1)^(1/3) - 2^(1/3)) or y = 6 (-2)^(1/3) ((-2)^(1/3) - 1)

Substitute back for x = y + 7:

Answer: x = 6 (2^(1/3) + 2^(2/3)) + 7 or x = 6 (-2)^(1/3) ((-1)^(1/3) - 2^(1/3)) + 7 or x = 6 (-2)^(1/3) ((-2)^(1/3) - 1) + 7

Answer 2

Even tho it’s not an awnser thx for helping