How do I factor this: x^3 -9x^2 +27x -27

Question

mr.math · Answer

Write $$x^3 -9x^2 +27x -27=(x^3-27)+(-9x^2+27x)=(x^3-3^3)-9x(x-3)=\dots$$

Does this help?

anonymous · Answer

let me try:
(x^3 - 3^3) - 9x (x-3)
1^3 (x-3) -9x (x-3)
(x-3) (-9x) 

lol am i right :S?

mr.math · Answer

Not quite. You first need to know the formula for the difference of two cubes: $$(a^3-b^3)=(a-b)(a^2+ab+b^2).$$ http://www.mathsisfun.com/algebra/polynomials-difference-two-cubes.html

mr.math · Answer

Now we will apply this to the problem in our hand. You wanna try?

anonymous · Answer

im so confused.. i havent learned this formula.. All i really needed from this equation was the zeros.. is there a easier way to find zeros than factoring?

mr.math · Answer

Don't be. It's an easy-to-learn formula. You can find help in understanding it in the link I gave above.
Now back to our problem (and by applying that formula), we can write:
$$(x^3 - 3^3) - 9x (x-3)=(x-3)(x^2+3x+9)-9x(x-3)$$
$$=(x-3)(x^2+3x+9-9x)=(x-3)(x^2-6x+9)$$
$$=(x-3)(x-3)^2=(x-3)^3.$$

mr.math · Answer

Makes sense?

anonymous · Answer

how did you get from: 
=(x−3)(x^2+3x+9−9x)
to
=(x−3)(x−3)^2

anonymous · Answer

oh no not that one sorry.

anonymous · Answer

(x^3−3^3)−9x(x−3) 
to
(x-3) (x^2+3x+9-9x)

anonymous · Answer

(2x+3)^_4

anonymous · Answer

??????///