Question

Theory of equation depressed cubic
@ganeshie8

Answer 1

If a , b , c are the roots of the equation x^3 -x -1 =0
then,
\[\frac{ 1+a }{ 1-a }+\frac{ 1+b }{ 1-b }+\frac{ 1+c }{ 1-c }\]
Has the value equal to what?

Answer 2

@mathslover

Answer 3

Should we consider the x^2 term as 0

Answer 4

So it would become
\[\huge a+b+c = 0\]
\[\huge ab+ac+bc = -1\]
\[\huge abc=1\]

Answer 5

Why would we consider x^2 term as zero? And is there any x^2 term?

Answer 6

nope it is a depressed one

Answer 7

Depressed one? Sorry, can you please explain.

Answer 8

Depressed means just when there is a term missing in a normal (any) polynomial
here x^2 term is missing so it is depressed

Answer 9

Oh, okay...

Answer 10

I am getting thousands of equations but not sure which to use i really need help
Or maybe i am making it too complicated somewhat

Answer 11

I'm not sure. May be @Hero @whpalmer4 @jim_thompson5910

Answer 12

do u have an answer for it ?
like the finl sol ?

Answer 13

Maybe this could be of any help:
if you just get it to commun denominator, in the first term you get something like this:
(1+a)(1-b)(1-c)+....
now, on the other side, since equation has coeficient besides x^2 equal to 0, this means that:
(x-a)(x-b)(x-c)=(x-a)(x^2-cx-bx+bc)=0
so putting this two facts together:
x^2-cx^2-bx^2=0
or
1-c=b

Answer 14

The answer is -7

Answer 15

@thomaster
And thank you @myko

Answer 16

@AkashdeepDeb Its of CBSE help me I am new to this board

Answer 17

i would try solve for x :P

Answer 18

so you would have a real , b,c complex lol
shall we try that ?

Answer 19

continuing with my start, :). Sincé the factors (x-a)(x-b)(x-c) can be put in any orther this means that 1-c=a, 1-a=c ... and so on

Answer 20

Let's seee but i doubt it would lead to something fruitful i already have around 150 sums to complete and most of them are from jee

Answer 21

Not that i doubt your mathematical ability

Answer 22

http://www.wolframalpha.com/input/?i=+x^3+-x+-1+%3D0

Answer 23

That's cardano's formula , that won't lead anywhere bu to our own doom

Answer 24

basicly you get 3 equalities:
1-c-b=0
1-c-a=0
1-a-b=0
from here b=a and b+a=1

Answer 25

so a=b=0.5

Answer 26

and c you get from any other relation

Answer 27

Substituting the values i am not getting -7 which is the correct answer
@myko

Answer 28

but the answer would be 9, not -7. So maybe I made a mistake somewhere...:(

Answer 29

It is comming -1.6666666666666666666................. :-(

Answer 30

\[\frac{1+a}{1-a} + \frac{1+b}{1-b} + \frac{1+c}{1-c}\]
\[ = \frac{(1+a)(1-b)(1-c) + (1+b)(1-a)(1-c) + (1+c)(1-a)(1-b)}{(1-a)(1-b)(1-c)}\]
\[ = \frac{3abc-ab-ac-bc-a-b-c+3}{-abc+ab+ac + bc -a-b-c+1}\]
\[=\frac{3abc - (ab+bc+ac) - (a+b+c) + 3}{-abc + (ab+ac+bc) - (a+b+c) + 1}\]

Now use the equations that you can get from the cubic equation to get your answer.

Answer 31

Okay ,
Are we supposed to include the x^2 term its coefficient being 0
and thank you akshay for being patient and explaing this. :)

Answer 32

Yes, the co-efficient of the \(x^2\) HAS to be zero because it is non-existent. So you can write it like this: 0.\(x^2\)

And my name is Akashdeep [or Akash] and not Akshay. :)

Answer 33

Ok AKASHDEEP thank you

Answer 34

I am wondering what is wrong woth my reasoning? Can`t see the mistake...

Answer 35

ok found it, :)

Answer 36

what was it

Answer 37

a+b+c=0

Answer 38

but anyway answr is 7 not -7

Answer 39

i got -7 now it is correct