Difference of Cubes: (2x-3)^3 - (x-6)^3

[Math_Junkie]

Can someone check my answer please?

(2x-3)^3 - (x-6)^3
= (2x-3-x+6)((2x-3)^2+(2x-3)(x-6)+(x-6)^2)
= (x+3)(4x^2-12x+9+2x^2-12x-3x+18+x^2-12x+36)
= (x+3)(7x^2-39x+63)

 

[Deleted member 4993]

Can someone check my answer please?

(2x-3)^3 - (x-6)^3
= (2x-3-x+6)((2x-3)^2+(2x-3)(x-6)+(x-6)^2)
= (x+3)(4x^2-12x+9+2x^2-12x-3x+18+x^2-12x+36)
= (x+3)(7x^2-39x+63)

Looks good to me...

In these problems, I do the multiplication twice - second time without looking at the first result and if possible taking adifferent direction (e.g. first time collect the square terms first - second time collect the constants first , etc.)