# cubics

• Feb 2nd 2010, 05:23 AM
usagi_killer
cubics
Prove that for a cubic of the form $\displaystyle a(x-b)^3+d$ then there exists at least one set of real solutions $\displaystyle \{a,b,d\}$ such that:

$\displaystyle y_1=a(x_1-b)^3+d$

$\displaystyle y_2=a(x_2-b)^3+d$

$\displaystyle y'_1 = 3a(x_1-b)^2$
• Feb 2nd 2010, 09:20 AM
HallsofIvy
Quote:

Originally Posted by usagi_killer
Prove that for a cubic of the form $\displaystyle a(x-b)^3+d$ then there exists at least one set of real solutions $\displaystyle \{a,b,d\}$ such that:

$\displaystyle y_1=a(x_1-b)^3+d$

$\displaystyle y_2=a(x_2-b)^3+d$

$\displaystyle y'_1 = 3a(x_1-b)^2$

Strictly speaking, what you want to prove here is NOT true! If $\displaystyle x_1= x_2$ and $\displaystyle y_1\ne y_2$, the first two conditions are incompatible. I am going to assume that $\displaystyle x_1\ne\x_2$.

Since $\displaystyle y'= 3a(x- b)^2$ that third condition follows from the first so we really have only two conditions. That means we are free to choose one of the parameters to be whatever we want. Choosing $\displaystyle b= x_1$ makes the first condition $\displaystyle y_1= d$. With $\displaystyle b= x_1$ and $\displaystyle d= y_1$, the second condition becomes $\displaystyle y_2= a(x_2- x_1)^3+ y_1$ which gives $\displaystyle a= \frac{y_2-y_1}{(x_2- x_1)^3}$ as long as $\displaystyle x_1\ne x_2$.

If we have $\displaystyle x_1= x_2$ and $\displaystyle y_1= y_2$, then we really only have one condition. Choose $\displaystyle b= x_1$ and $\displaystyle d= y_1$. a now can be any number at all.
• Feb 2nd 2010, 07:06 PM
usagi_killer
Sorry $\displaystyle x_1 \neq x_2$ and $\displaystyle y_1 \neq y_2$
• Feb 2nd 2010, 08:25 PM
Pulock2009
hey this comes under theory of equations,right??? i expanded (x-b)^3 and got a cubic. if the roots be y1,y2,y3 then i got the following:
y1+y2+y3=-3b
y1y2+y2y3+y3y1=-3b^2
y1y2y3=d/a-b^3

as evident there should be 3 roots but i simply couldnot find y1,y2,y3.there seems to be some kind of a proof not by usual methods.