Factorize $(x-3)^2+(x-9)^3=2(x-6)^3$

Question

How to factorize : $$(x-3)^2+(x-9)^3=2(x-6)^3$$

I came across this in a Vedic Maths chapter. Using Vedic Maths, we get the value of x as 6 but I'm unable to factorize this using tradional methods.

I expanded this to : $$(x^2-6x+9)+(x^3-27x(x-9)-729) = 2[x^3-18x(x-6)-216]$$ then after expanding this, $$x^3-10x^2-21x+288=0$$ which I'm unable to factorize.

Can anyone please tell me how to proceed further?

Answer 1

What you posted doesn't factor nicely, but (assuming a typo somewhere) this one does: $$(x-3)^{\color{red}{3}}+(x-9)^3-2(x-6)^3 = 54 (x - 6)$$

Answer 2

Assuming, as dxiv did, the typo, let us exploit the symmetry and let $x=y+6$ which makes $$(x-3)^{3}+(x-9)^3-2(x-6)^3 = 2 y \left(y^2+27\right)-2y^3=54y=54(x-6)$$