use the quadratic equation to solve for x^3-216=0

factor as the difference of two cubes. you have \[x^3-6^3\]

factors as \[(x-6)(x^2+6x+36)\]

so one solution to \[(x-6)(x^2+6x+36)=0\] is \[x=6\]

the other two come from the quadratic \[x^2+6x+36=0\] and the solutions will be complex. do you need these as well?

i used \[a^3-b^3=(a-b)(a^2+ab+b^2)\] to factor

if you want to solve \[x^2+6x+36=0\] \[x^2+6x=-36\] \[(x+3)^2=-36+3^2=-27\] \[x+3=\pm\sqrt{-27}\]

the formula is good... but any case u can see dividing and obtain the other part if we know one factor

and so \[x=-3\pm3\sqrt{3}i\]

so after you get the x=6 you apply in for the second step?

you could also do it the quick way \[6e^{i\cdot0},6e^{i\cdot2\pi/3},6e^{i\cdot4\pi/3}\]

;)

yeah sure

not how they wanted it ...but it is quicker

I asked because when i used the a^3 - b^3 i don't get the x^2 + 6x+36

or the real quick way, say it is \[6\times 1, 6\times \frac{-1}{2}\pm\frac{\sqrt{3}}{2}i\]

ok lets go slow

what you are supposed to recognize is "the difference of two cubes" \[a^3-b^3\]

zarkon I feel like you are speaking another language lol I have no idea what those are, i HAVEN'T LEARNED THAT YET.

the difference of two cubes always factors, as does the sum of two cubes. the check is by multipication

Okay I got that part satellite73

that is ok, but it is clear what it means right? "difference" means subtraction and "cubes" means something cubed

yes

and it is a fact that \[a^3-b^3=(a-b)(a^2+ab+b^2)\]

if you multiply out on the right, you will get the expression on the left

now you are given \[x^3-216\] and since the first term is a cube, figure that the second one must be a cube also

the first side is x-6

and check that \[6^3=216\] so you have \[x^3-6^3\] which looks just like \[a^3-b^3\] right?

ohh okay

so i write \[x^3-6^3=(x-6)(x^2+6x+6^2)\] just like i would if they were \[a^3-b^3\]

could it also be written as (x-6)(x+6)(x-6)

absolutely not

there is only one way to factor an expression, just like there is only one way to factor a number

if you multiply out what you wrote, you get \[x^3-6 x^2-36 x+216\] not \[x^3-216\]

try it and you will see

what if I used (x+6)(x+6) to get the second part?

or is it that whatevever the cube equals I must apply it in the second? for example x=6. so therefore I could just place the 6 in the second.

ok at the risk of sounding like a math teacher (i apologize in advance) there is only one way to factor the difference of two cubes. so \[x^3-216=(x-6)(x^2+6x+36\] and no other.

the second part \[x^2+6x+36\] does not factor using integers

you cannot do it. that is why the zeros are complex numbers.

It's okay the way you sounded but I needed it in a simple form sorry to sound slow but yea my brain shut down on me tonight.

so its just something I must know then? from what you did i noticed the 6 is applied in the b and the c^2. is it like that all the time? even my book did the same thing.

i guessed it because the first term was a cube. it was \[x^3\]

so the only way this could factor as the difference of two cubes is if the second term was a cube as well

now not being a calculating genius, i took out a calculator and put in \[\sqrt[3]{216}\] and out popped 6, just like i suspected

that is how i knew it was \[x^3-6^3\]

so if i see \[x^3-8\] i think "x cubed minus 2 cubed" and write \[x^3-8=(x-2)(x^2+2x+4)\]

\[x^3-27=(x-3)(x^2+3x+9)\] and so one

right and the 2 and 3 were applied in the b and c

exactly!

fantastic, you're a life saver!!!! lol

one more thing. if you have "the SUM of two cubes it factors as well \[a^3+b^3=(a+b)(a^2-ab+b^2)\]

notice the difference in the placement of the minus sign

that equation def came in handy

are you good with math?

fair to middlin'

lol okay, thank you so much, I appreciate it

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