let cube root of 1 be x.
=> x^3 = 1
=> x^3 - 1 = 0
=> (x-1) (x^2 + x + 1) = 0
=> x=1 or x^2+ x+1=0
=> x=1 or x= [1+ i(3)^1/2 ] / 2 or [1- i(3)^1/2 ] / 2
where i is iota, (-1)^1/2
Hence cube root of one has three roots, one real and 2 imaginary. these imaginary roots are called w (omega) and omega square.
1+w +w^2 =0
1* w * w^2 = 1
2006-11-26 23:00:37
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answer #1
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answered by Manjari 2
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1
2006-11-27 00:04:25
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answer #2
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answered by Slimm D 3
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1
2006-11-26 22:42:07
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answer #3
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answered by ? 2
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The cube root of 1 is 1 itself
2006-11-26 22:40:35
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answer #4
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answered by Sankalp P 1
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(–1 ± i√3)/2 are cube roots of 1
2006-11-26 22:51:33
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answer #5
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answered by Prasad.S 2
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The cube root of 1 is 1. 1 cubed is 1x1x1=1
2006-11-26 22:55:30
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answer #6
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answered by Max 6
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1, (i√3-1)/2, and (-i√3-1)/2
2006-11-26 22:49:20
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answer #7
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answered by Pascal 7
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(dice root of 25) over 5. 'least confusing kind' skill getting the novel out of the denominator. a million/5 ^a million/3 . Multiply a million/5 with the help of 25/25 so as that the denominator is a proper dice. That makes the numerator equivalent to twenty-5.
2016-11-27 01:11:03
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answer #8
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answered by ? 3
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It is just think of the number that when you multi[ply to itself 3 times teh answer is 1 and that is 1
2006-11-26 23:11:31
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answer #9
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answered by jdash01 3
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Its '1'
As one / one 3 times is 1.
I hope its clear to you.
2006-11-26 22:42:30
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answer #10
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answered by Mayank 2
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