If x=1+i is a root of the equation x^3 −ix+1−i=0 , then the other real root is


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Question Number 49460 by Pk1167156@gmail.com last updated on 07/Dec/18

$If x = 1 + i is a root of the equation$ $x^{3} - i x + 1 - i = 0, then the other real$ $root is$
Commented by Kunal12588 last updated on 07/Dec/18

$i d o i t w i t h n o r m a l r e a l w a y . s o i t$ $s h o u l d b e w r o n g .$ $x^{3} - i x + 1 - i = (x - 1 - i) (x + i) (x + 1)$ $r o o t s a r e$ $x_{1} = 1 + i, x_{2} = - i, x_{3} = - 1 . s o r e a l r o o t = - 1 .$ $p l e a s e i n f o r m i f i t i s w r o n g .$
Commented by Abdo msup. last updated on 08/Dec/18

$(x - 1 - i) (x + i) (x + 1) = (x - 1 - i) (x^{2} + x + i x + i)$ $= (x - 1 - i) (x^{2} + (1 + i) x + i)$ $= x^{3} + (1 + i) x^{2} + i x - x^{2} - (1 + i) x - i - {i x}^{2} - i (1 + i) x + 1$ $x^{3} - x - i x + x + 1 - i = x^{3} - i x + 1 - i s o y o u r a n s w e r i s$ $c o r r e c t s i r .$
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