Sum and products of roots

[Dean54321]

If one root of the equation x³-bx²+cx-d=0 is equal to the product of the other two, show that (c+d)²=d(b+1)²

Therefore the roots are: a, B, and aB

I worked out 3 equations (though they could be wrong):
1. a²b² = d
2. a +B + aB = b
3. c = aB(1+B+a)

I expanded the LHS and RHS and got the answer, I was wondering if there was a more elegant way.

 

[Zermelo]

Hey man, google Vieta’s formulas, those give you a releationship between the coefficients and roots of a polynomial. I think those should help. For example, if x y z are the roots, one of the formulas tells you that -(-d) = xyz, that means d =x^2, because one root is the product of other two.

 

[lex]

Can't think of anything particularly elegant.

[MATH]\alpha \text{, } \beta \text{, } \gamma\hspace2ex \text{three roots}\\ \text{ }\\ \beta\gamma +\beta +\gamma =b\\ \beta^2\gamma + \gamma^2\beta+\beta\gamma=c\\ (\beta \gamma)^2=d[/MATH]put
u=[MATH]\beta \gamma[/MATH]v=[MATH] \beta + \gamma[/MATH]
[MATH]u+v=b \hspace2ex[/MATH] (1)
[MATH]uv=c-u\hspace2ex[/MATH] (2)
[MATH]u^2=d \hspace2ex[/MATH] (3)

Multiply (1) by u:
(1)' [MATH]u^2+uv=bu[/MATH]substitute (2), (3) into (1)'

[MATH]d+c-u=bu\\ u(b+1)=c+d[/MATH]squaring: [MATH]d(b+1)^2=(c+d)^2[/MATH]