Question

please solve
x^3-8=0

Answer 1

add 8 to both sides....

Answer 2

and then multiply by 1/3 exponential 1/3

Answer 3

then rewrite the right side in radical form and take the cube root of 8

Answer 4

3 should be written outside the sq root and since it's just 1 you don't have to write 8 to the first power

Answer 5

and ans is 2???

Answer 6

yes plug it back into the original equation.

Answer 7

you know there will be 3 roots of x ?
you only need integer one ?

Answer 8

I need to plug my pc back in brb

Answer 9

why can't we use formula a^3-b^3=(a-b)(a^2+ab+b^2)????

Answer 10

you should use that formula...

Answer 11

if we use that formula we will get one real and two complex roots

Answer 12

correct

Answer 13

thats what i want to ask....
x^3=8
x^3=2^3
x=2
give only real root
but applying formula give 1 real and 2 complex roots
why this difference come on applying two different(but correct) methods??????

Answer 14

"x^3=2^3
x=2 "

this is correct but incomplete.

x^3=2^2
x^3-2^3=0
now apply a^3 - b^3 formula
this method is correct and complete.

Answer 15

\[x^3 -8 \\ x=\sqrt[3]{8}\\ x=\sqrt[3]{2^3}\\ x = 2\]

Answer 16

I shall join the long line of people answering.
\(\large{x^3-8=0}\)
\(\large{x^3=8}\)
\(\large{x=\sqrt[3]{8}}\)
\(\large{x=2}\)

I feel happy now.

Answer 17

Go for hartnn's explanation, he did the complexes as well

Answer 18

lol there are 3 solutions!
listen to @hartnn

Answer 19

everyone else stop saying 2 is the answer...

Answer 20

what is this, a "who draws better latex" competition ?!

Answer 21

Let p(x)=x^3-8
Since you've known x=2 is a root
Then (x-2) is a factor of p(x)
Do a synthetic division

Then use the quotient and use any factoring methods to get your complex roots

Answer 22

\(\Huge{\color{red}{\text{Y}}}\Large{\color{blue}{\text{E}}}\LARGE{\color{green}{\text{S}}}\)
@Luis_Rivera

Answer 23

lol

Answer 24

Do you understand rosy?

Answer 25

\[a^3-b^3=(a-b)(a^2+ab+b^2)\\x^3-2^3=0\\(x-2)(x^2+2x+4)=0\]now solve these\[x=2\\and\\(x+1)^2=-3\\x=\pm(i\sqrt{3}-1)\]

Answer 26

ok back to vacation in paris:) see ya suckers

Answer 27

Hey mind your language zzr0ck3r :)

Answer 28

by the way, @Z3R0, i wonder what problem you're working on ? cause it certanly aint this one !

Answer 29

These are the basic steps for beginners like him

Answer 30

diving in!
(x-2) (x^2+2 x+4) = 0