Question

If a/b + b/a = -1. Then find a^3 + b^3.

Answer 1

here..we easily see a^2 + b^2 = ab
now a^3 + b^3 = (a+b) (a^2+ b^2 -ab)
so the ans is 0!!

Answer 2

wouldnt:\[a^2+b^2 = -ab\]?

Answer 3

oh yes am sorry!! leme figure out something else..

Answer 4

well in that case you see a^3 - b^3 = 0
thus a=b
now should be easier..

Answer 5

So a^3 + b^3 = a^ + a^3
= 2a^3.
But does that help?

Answer 6

well i dont guess it can be simplified further..

Answer 7

no no.wait,,theres something fishy...a=b does not satisfy a/b + b/a = -1
hmmn..

Answer 8

which means a and b are complex..

Answer 9

i dont think there are any solutions to that equation. if you try to solve:\[x+\frac{1}{x}=-1\]You end up with:\[x^2+x+1=0\]which only have complex solutions.

Answer 10

yeah, a and b would have to be complex.

Answer 11

do this..let a/b =x
so you have a quadratic eqn x^2+ x + 1 =0
thus you solve a/b = something,,so,,a = b(something)
try and make that substitution..dont know if this'd be a great help//

Answer 12

in fact, i think:\[\frac{a}{b}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\]and :\[\frac{b}{a}=-\frac{1}{2}-i\frac{\sqrt{3}}{2}\]works.

Answer 13

from (a+b)(a^2 + b^2 -ab) ,,factor like this :
a(1 + b/a) * (a^2)(1 + (b/a)^2 - b/a )
now try.. hmmn..

Answer 14

note that b^2 = a and a^2 =b

Answer 15

yes thats right.

Answer 16

first we get a^2 +b^2 = -ab now putting this in a^3 +b^3 we get
a+b*(-2ab) then using (a+b)^2.... a+b=\[\sqrt{ab}\]...so ans. should be -2[ab\[\sqrt{?}\]]

Answer 17

sqrt(ab)..*......nt "?"