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Mathematics
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (zarkon):

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

OpenStudy (zarkon):

;)

OpenStudy (anonymous):

yeah sure

OpenStudy (zarkon):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok lets go slow

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Okay I got that part satellite73

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

the first side is x-6

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohh okay

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

absolutely not

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

try it and you will see

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

exactly!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

notice the difference in the placement of the minus sign

OpenStudy (anonymous):

that equation def came in handy

OpenStudy (anonymous):

are you good with math?

OpenStudy (anonymous):

fair to middlin'

OpenStudy (anonymous):

lol okay, thank you so much, I appreciate it

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