Question

proof of 3 = 0

Answer 1

x^2+x+1=0
x^2=-1-x
x=-1-1/x
x^2+x+1
x^2-1-1/x + 1
x^2-1/x = 0
x^3-1 = 0
x^3 = 1
x = 1
1^2+1+1=0
3=0

Answer 2

i cant understand how you get the 4th line

Answer 3

in starting i took x^2+x+1 = 0
in thirf line i got x=-1-1/x
i again wrote 4th line i had not derived it it is just taken up from above for further solving

Answer 4

u shud win noble prize
from my side u r done

Answer 5

i think the substitution is wrong

Answer 6

the value of x is imaginary number???

Answer 7

kumarsajan is 100% correct

Answer 8

of course it is imaginary

Answer 9

so the equation has no answer at all, every step is incorrect because how can you devide by an imaginary number? and how can you multiply by an imaginary number, so it will gige you an imaginary answer.

:))

Answer 10

@kumarsajan
x is imaginary and is taken equal to 1 which is real, nice ans by u

Answer 11

should I give it away?

Answer 12

yea

Answer 13

yes. ofcourse

Answer 14

\[x^3-1=(x-1)(x^2+x+1)\]

Answer 15

Yes, exactly. What this derivation above shows is that

x^2 + x + 1 = 0 ----(*)

implies that x^3 - 1 =0

But x^3 - 1 =0 means that (x-1)( x^2 + x + 1 ) = 0

i.e., x - 1 = 0 OR x^2 + x + 1 = 0

[[ Logically this is the same as writing (A => A or B).
Unless now (A => B), the fact that (A => A or B) will NOT mean that B is true. ]]

Now x = 1 does in fact not satisfy (*).
Hence it is not true that x- 1 =0 and all we have shown that

x^2 + x + 1 = 0 ==> x^2 + x + 1 = 0

That's the logic of the situation.
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There's also something interesting going on here. If you choose one of the complex roots of z^3 - 1 =0, i.e.,
\[z = \frac{-1 \pm \sqrt{3}}{2}\]
then in fact z^2 + z + 1 = 0. Which is of course completely consistent.

Answer 16

@kumarsajan, you should not trivialize "imaginary" number. As you know division, multiplication, and other mathematical operations are indeed possible with complex and imaginary numbers. If your studies take you into electronics, you will soon find that many calculations will involve imaginary numbers. They use "j" rather than "i" as "i" is the symbol for current. The j operator is used in calculating impedance when inductive or capacitance reactance is involved.

Answer 17

Good work to radar , Jamesj and Zarkon
Kumar lost

no problem good work to all