Question

what is the simplest form of

Answer 1

my answer is 9

Answer 2

-28 -18 9 or 57

Answer 3

if I am wrong can someone explain it ?

Answer 4

your answer choices don't include -9?

Answer 5

it might be -28 cause I get - 83/3

Answer 6

what are your anwser choices

Answer 7

answer

Answer 8

exponentiation binds tighter than unary minus, so that's effectively

-1*27^(2/3)

not
(-27)^(2/3)

Answer 9

9 is correct.

Answer 10

-28 or -18 or 9 or57

Answer 11

http://www.wolframalpha.com/input/?i=-27%5E%282%2F3%29&lk=4&num=1

Answer 12

The website that I use to check my answers is:
http://www.algebra.com/services/rendering/simplifier.mpl

Answer 13

so try -27^(2/3) there :-)

Answer 14

Ohh I didn't notice that it was -27, I read it as positive, sorry for the confusion.

Answer 15

anything with a negative number in the radical produces an imaginary number... i

Answer 16

so... ughhh let's just twiddla this thing

Answer 17

http://www.twiddla.com/1469553

Answer 18

that's true, but the question here is did they intend to write the negative sign in the radical, or were they simply sloppy. what they wrote is the equivalent of -1*27^(2/3)

Answer 19

I got 9

Answer 20

IF AND ONLY IF THERE IS A () in the radical

Answer 21

if not then it would've be cube root of -729 and that's 9i

Answer 22

It just so happens that -27 is a perfect cube: the cube of -3.

However, if you write -27^(2/3), the exponentiation applies only to the 27, not to the -27. Thus, you are evaluating -[27^(2/3)].

Note that the cube root of 27 is 3. Therefore, as I see it, 27^(1/3) = 3, and 27^(2/3)=(3)^2, or 9.

And that 9 is inside brackets: [9].

And you must write that negative sign (from the 3rd line of type, above) in front of that [9].

That's my take on this problem.

Answer 23

Too bad that the four possible answers you've shared apparently don't include -9.

U-Doll: note that we're discussing third roots here, not square roots, so that discussion about i is irrelevant, isn't it?

Answer 24

dude I said cube root... I didn't mention a square did I?

Answer 25

and get a grip any radical with a negative sign inside produces an imaginary number... read what whpalmer said

Answer 26

what does the question really tell us? was it their sloppiness or something else?

Answer 27

pfft forget your explanation.. the answer is 9. how to check... well 9^3 which is 9 X 9 X 9 = 729

Answer 28

UsukiDoll: quoting you:

"if not then it would've be cube root of -729 and that's 9i":

Please justify your statement.

Notice that (9i)^2=-81, and so (9i)^3=(-81)(9i) = -729i (not the same as -729).

The cube root of 729 is 9; the cube root of (-729) is -9.

Answer 29

omg don't teach me on roots. I know what they are!

Answer 30

If you believe this is supposed to be the 2/3 root of -27,

(-27)^(2/3) = (-1*27)^(2/3) = (-1)^(2/3) * (27)^(2/3)
-1^(2/3) is complex.

Answer 31

Have it your own way, UD. I'm leaving.

Answer 32

Here is what I wrote earlier...IF YOU WANT TO GO BACK AND FORTH !

"I got 9 ... IF AND ONLY IF THERE IS A () in the radical...if not then it would've be cube root of -729 and that's 9i"

Answer 33

yeah please leave. you don't have a clue

Answer 34

Thank you for your kind compliments.

Answer 35

THEN GO!

Answer 36

*kicks a can* bah! Thinking that I don't know cube roots. YOU ARE SADLY MISTAKEN MISTER!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 37

I wasn't even saying square roots and this guy went all over wt fdkslfjsdafjdsklfjdl;jfl;fkl;sajfdsajfkdsa

Answer 38

cube root of -729 is not 9i

the 3 cube roots of -729 are -9, 9e^(-i*pi/3), and 9e^(i*pi/3)

9i*9i*9i = 729 * i^3 = 729 * i^2 * i = 729 * (-1) * i = -729i

-729i != -729

therefore, 9i is not a cube root of -729

Answer 39

i is an imaginary number right? ... how would you explain the square root of -4?

Answer 40

the way I was taught.. sqrt(-4) is 2i

Answer 41

i^2 is a negative 1...I still remember that

Answer 42

yes. sqrt(-4) = 2i
sqrt(-4) = sqrt(-1*4) = sqrt(-1)*sqrt(4) = i*2 = 2i

Answer 43

but the numeric forms of the cube root of -729 are -9 and two complex conjugates:
4.5 + 7.79423i and 4.5 - 7.79423i

(approximately)

let a = 4.5
b = 7.79423i

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

plug in the numbers:

(4.5)^3 + 3(4.5)^2(7.79423i) + 3(4.5)(7.79423i)^2 + (7.79423i)^3
i^2 = -1
(4.5)^3 + 3(4.5)^2(7.79423i) - 3(4.5)(7.79423)^2 - (7.79423)^3*i
91.125 + 473.5i - 820.125 - 473.5i
91.125-820.125
-729