anonymous
  • anonymous
please solve x^3-8=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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UsukiDoll
  • UsukiDoll
add 8 to both sides....
UsukiDoll
  • UsukiDoll
and then multiply by 1/3 exponential 1/3
UsukiDoll
  • UsukiDoll
then rewrite the right side in radical form and take the cube root of 8

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More answers

UsukiDoll
  • UsukiDoll
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
anonymous
  • anonymous
and ans is 2???
UsukiDoll
  • UsukiDoll
yes plug it back into the original equation.
hartnn
  • hartnn
you know there will be 3 roots of x ? you only need integer one ?
UsukiDoll
  • UsukiDoll
I need to plug my pc back in brb
anonymous
  • anonymous
why can't we use formula a^3-b^3=(a-b)(a^2+ab+b^2)????
hartnn
  • hartnn
you should use that formula...
anonymous
  • anonymous
if we use that formula we will get one real and two complex roots
hartnn
  • hartnn
correct
anonymous
  • anonymous
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??????
hartnn
  • hartnn
"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.
anonymous
  • anonymous
\[x^3 -8 \\ x=\sqrt[3]{8}\\ x=\sqrt[3]{2^3}\\ x = 2\]
austinL
  • austinL
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.
anonymous
  • anonymous
Go for hartnn's explanation, he did the complexes as well
zzr0ck3r
  • zzr0ck3r
lol there are 3 solutions! listen to @hartnn
zzr0ck3r
  • zzr0ck3r
everyone else stop saying 2 is the answer...
anonymous
  • anonymous
what is this, a "who draws better latex" competition ?!
anonymous
  • anonymous
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 |dw:1378828971391:dw| Then use the quotient and use any factoring methods to get your complex roots
austinL
  • austinL
\(\Huge{\color{red}{\text{Y}}}\Large{\color{blue}{\text{E}}}\LARGE{\color{green}{\text{S}}}\) @Luis_Rivera
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
lol
anonymous
  • anonymous
Do you understand rosy?
zzr0ck3r
  • zzr0ck3r
\[a^3-b^3=(a-b)(a^2+ab+b^2)\\x^3-2^3=0\\(x-2)(x^2+2x+4)=0\]now solve these\[x=2\\and\\(x+1)^2=-3\\x=\pm(i\sqrt{3}-1)\]
zzr0ck3r
  • zzr0ck3r
ok back to vacation in paris:) see ya suckers
anonymous
  • anonymous
Hey mind your language zzr0ck3r :)
anonymous
  • anonymous
by the way, @Z3R0, i wonder what problem you're working on ? cause it certanly aint this one !
anonymous
  • anonymous
These are the basic steps for beginners like him
nincompoop
  • nincompoop
diving in! (x-2) (x^2+2 x+4) = 0

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