This solution is from Arun Iyer, SIA High School and Junior College.

The given equation is x ^3/2 - 8x ^-3/2 = 7.

Multiply throughout by x ^3/2 and rearranging we get

(x ^3/2) ² - 7x ^3/2 - 8 = 0

This can be factorised as:

(x ^3/2 - 8)(x ^3/2 + 1) = 0

Case 1:

Consider,

x ^3/2 + 1 = 0 or x ^3/2 = -1

Squaring both sides, x ³ = 1 so the three cube roots of unity are solutions.

Now (for k=0,1,2,3...) therefore .

Putting k=0,1,2 we get:

x = e ⁰ (which is 1) or .

Case 2:

Consider,

x ^3/2 - 8 = 0 or x ^3/2 = 8

Squaring both sides, x ³ = 64 so the three cube roots of 64 are solutions.

Now (for k=0,1,2,3...) therefore .

Putting k=0,1,2 we get: x = 4 or .

So the six roots are:

• x ₁ = 1
• x ₂ =
• x ₃ =
• x ₄ = 4
• x ₅ =
• x ₆ =

Substituting x=1 into the given equation we need to recognise that 1 ^3/2 has two values +1 and -1 so that whereas x ^3/2 = 1 does not satisfy the equation the other value x ^3/2 = -1 does satisfy it and hence x=1 is a solution of the equation.