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Solve equation for real solution x^3-1=0

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indiraa | Student, Undergraduate | eNoter

Posted December 12, 2011 at 1:45 AM via web

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Solve equation for real solution x^3-1=0

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baracuda123 | Student, Grade 10 | eNotes Newbie

Posted December 12, 2011 at 3:53 AM (Answer #1)

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or...

x^3-1=0

you just add 1 to both sides

x^3= 1

then you know the cube root of 1 is just 1

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 12, 2011 at 2:22 AM (Answer #2)

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You need to remember the formula of the difference of two perfect cubes:

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

Comparing the difference of two cubes to the given equation yields:

`x^3-1 = (x-1)(x^2 + x + 1)`

You need to solve for x the product `(x-1)(x^2 + x + 1)`  = 0.

x - 1 = 0 => x = 1

Notice that `x^2 + x + 1`  > 0 `AA` x `in`  R

The real solution to the given equation is x = 1.

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ainsleyharris | Student, Grade 9 | Honors

Posted December 21, 2011 at 10:03 AM (Answer #3)

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work backwards, 0+1 =1. therefore, x^3 has to equal 1, meaning that x is 1.

1^3=1. 1-1=0.

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atyourservice | TA , Grade 10 | Valedictorian

Posted April 16, 2014 at 8:25 PM (Answer #4)

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x^3-1=0

      +1 +1

x^3 =1

and the answer is 1 because cube root of 1 is 1

1x1x1= 1

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