Question

x^2-ax-1=0, x^3+1/x^2+x^2-1/x^3+1 what is a?

Answer 1

I asked wolfram but still i don't get its answer.

Answer 2

x^3+1/x^2+x^2-1/x^3+1 =???
you have 2 unknowns so dont you need 2 equations to solve it?

Answer 3

ooo... shiny, cheers

Answer 4

wait sorry i was trying to say put this equation in terms of a and do factoring

Answer 5

ah, k, in that case:
x^2-ax-1=0
-1 = x^2-ax
a= x- (1/x)

Answer 6

hmmm. What do you mean by k?

Answer 7

so... shortened version of okay, not a constant

Answer 8

sorry

Answer 9

ok

Answer 10

(x^3+1)/(x^2+x^2-1)/(x^3+1) is this eqn 2?
x^3+(1/x^2)+x^2-(1/x^3)+1 ...or this?

Answer 11

so should i make it like (x-1)(x^2+1+1/x^2)(x^+1+1/x^1)

Answer 12

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

so a^2= x^2+(1/x^2) +2

therefore take that out of above leaves remainder:
x^3-(1/x^3)-1

Answer 13

so now u have:
a^2 + x^3-(1/x^3)-1
so just keep going untill you get

=a^+ 3a +a^3 and remainder of +3

Answer 14

so reduces to =a^2+ 3a +a^3+3

Answer 15

oh thanks.