The nature of the roots of the equation (x²–6x+9)² = x³–4x²+9x is

the given equation is (x²–6x+9)² = x³–4x²+9x 

${\left({\mathrm{x}}^2-6\mathrm{x}+9\right)}^2=\mathrm{x}\left({\mathrm{x}}^2-6\mathrm{x}+9\right)+2{\mathrm{x}}^2$

Take x² – 6x + 9 = y.

Then y² – xy – 2x² = 0 y = 2x, –x

${\mathrm{x}}^2-6\mathrm{x}+9=2\mathrm{x}\Rightarrow {\mathrm{x}}^2-8\mathrm{x}+9=0$ Roots real and distinct and ${\mathrm{x}}^2-6\mathrm{x}+9=-\mathrm{x}\Rightarrow {\mathrm{x}}^2-5\mathrm{x}+9=0$

Roots are conjugate complex numbers.