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 1 = e^2kpi (for k=0,1,2,3...) therefore x = e^{2kpi /3} .

Putting k=0,1,2 we get:

x = e ⁰ (which is 1) or x = exp(±(2pi /3)) .

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 x³ = 64 e^2kpi (for k=0,1,2,3...) therefore x = 64^1/3 e^2kpi/3 .

Putting k=0,1,2 we get: x = 4 or x = 4exp(±2pi /3) .

So the six roots are:

• x ₁ = 1
• x ₂ = exp(2pi /3) = cos(2p/3) + isin(2p/3) = -1/2 + i Ö3/2
• x ₃ = exp(-2pi /3) = cos(-2p/3) + isin(-2p/3) = -1/2 - i Ö3/2
• x ₄ = 4
• x ₅ = 4exp(2pi /3) = 4cos(2p/3) + 4isin(2p/3) = -2 + 2i Ö3
• x ₆ = 4exp(-2pi /3) = 4cos(-2p/3) + 4isin(-2p/3) = -2 - 2i Ö3

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.