How do you factor $(x-3)^3 + (3x-2)^3$ ?

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Answer 1 · 2018-03-19T17:51:20

$=7(4x-5)(x^2-x+1)$

Both terms are cubes and the expression can be factored as:

$a^3 +b^3 = (a+b)(a^2 -ab+b^2)$

To make the process easier to follow,

let $(x-3) = m and (3x-2) =n$

$m^3 +n^3$

$= (m+n)(m^2-mn+n^2)$

$=(x-3+3x-2)((x-3)^2-(x-3)(3x-2) +(3x-2)^2)$

Simplify:

$=(4x-5)(x^2-6x+9 -(3x^2-11x+6)+9x^2-12x+4)$

$=(4x-5)(x^2-6x+9 -3x^2+11x-6+9x^2-12x+4)$

$=(4x-5)(7x^2-7x+7)$

$=7(4x-5)(x^2-x+1)$

How do you factor (x-3)^3 + (3x-2)^3(x-3)^3 + (3x-2)^3?