Factoring????

Question

How do I factor out:

(x+3)exponent of 3 negative (x-3) exponent of 3???

Could you please explaine the process!

Answer 1

so you have...
[(x+3)^3 - (x-3)^3]
right?

Use the difference of cubes formula...
x^2 - y^2 = (x-y)(x^2 + xy + y^2)

or, in our case...
[(x+3) - (x - 3)][(x+3)^2 + (x+3)(x-3) + (x-3)^2]

Simplifying...
[(x+3) - (x - 3)][(x+3)^2 + (x+3)(x-3) + (x-3)^2]
(6)[(x^2 + 6x + 9) + (x^2 - 9) + (x^2 - 6x + 9)]
(6)(3x^2 + 9)
(6)(3)(x^2 + 3)
18(x^2 + 3)

Answer 2

Not sure from your wording, but do you mean (x + 3)^3 - (x -3)^3?

If so, the you could recognize that you have the difference of two cubes and could use the formula A^3 - B^3 = (A-B)(A^2 + AB +B^2) where A = x+3 and B = x-3.

Here are the steps:

(x + 3)^3 - (x -3)^3 = [(x+3) - (x-3)][(x+3)^2 + (x+3)(x-3) + (x-3)^2]

if you need to completely simplify as well,

. . . .=[6][(x^2 + 6x +9) +(x^2 - 9) + (x^2 -6x +9)]

. . . .=[6][3x^2 +9]

. . . .=(18)(x^2 +3)

Answer 3

i believe you factor the first 2 parts first like

(x+3)(x+3)(x+3)
(x-squared+3x+3x+9)(x+3)
(x-squared+6x+9)(x+3)
and then i think you factor out the rest!

you multiply the first part then the outter part then the inner then the last! FOIL really simple.
you should get

(x-thirdpower+6x-squared+9x+3x-squared+18x+27) Group them
(x-thirdpower+9x-squared+27x+27)