Nasty equation help

(b) (i) Solve the equation y² + 8 = 9y.

(ii) Hence solve the equation x³ + 8 = 9x^3/2 .

I've managed to do part (i) but how on Earth do you do part (ii)? I've never been taught how to factorise or solve equations with fractional powers.

(b) (i) Solve the equation y² + 8 = 9y.

(ii) Hence solve the equation x³ + 8 = 9x^3/2 .

I've managed to do part (i) but how on Earth do you do part (ii)? I've never been taught how to factorise or solve equations with fractional powers.

(b) (i) Solve the equation y² + 8 = 9y.

(ii) Hence solve the equation x³ + 8 = 9x^3/2 .

I've managed to do part (i) but how on Earth do you do part (ii)? I've never been taught how to factorise or solve equations with fractional powers.

Part (i) is just a quadratic and you should be spotting the similarity between those two parts, look at how the numbers join up.

In fact, if you use the substitution $u = x^{3/2}$ you'll get $u^2 + 8 = 9u$ which is precisely what you solved in the first part; so you need only back-substitute now.

Hint: Let $y=x^{3/2}$.

EDIT: Too slow.

You can see in equation (ii) that 3/2 * 2 = 3. I.e., one term has double the exponent of the other, hence the similarity to equation (i)

Oooh okay thanks, i'll give it go