# How do we solve this cubic equation

• chwala

Gold Member

## Homework Statement

solve ## x^2/y + y^2/x=9, 1/x+1/y=3/4##[/B]

## The Attempt at a Solution

##9x^3+9y^3=81xy##
## 4x+4y-3xy=0##
##x^3+y^3-9xy=0, 12x+12y-9xy=0##
##x^3-12x+y^3-12y=0## [/B]

I would solve one equation for one variable and plug it into the other to see if that leads to a solution.

The equation you derived is interesting but it doesn't tell you what x and y are.

There is a trick to solving this particular problem that isn't (I don't think) relevant to other cubic equations.

See if you can factor ##\frac{x^2}{y} + \frac{y^2}{x}## in the following form: ##(\frac{1}{x} + \frac{1}{y})(Ax^2 + Bxy + Cy^2)##

• mfb
chwala said:

## Homework Statement

solve ## x^2/y + y^2/x=9, 1/x+1/y=3/4##[/B]

## The Attempt at a Solution

##9x^3+9y^3=81xy##
## 4x+4y-3xy=0##
##x^3+y^3-9xy=0, 12x+12y-9xy=0##
##x^3-12x+y^3-12y=0## [/B]

Solve ##4x+4y - 3xy = 0## for ##y## as a function of ##x##; that is, ##y = Y(x).## Substitute ##y= Y(x)## into the equation ##x^3+y^3 - 9xy = 0## to obtain the equation ##F(x) = 0,## where ##F(x) = x^3 + Y^3(x) - 9 x Y(x)##. Rewrite ##F(x)## as a rational function of ##x## and go on from there.

• chwala
ehild said:
So you have
##(x+y)(x^2-xy+y^2) = 9xy##

That was my hint, except in the alternative form:
##(\frac{1}{x}+\frac{1}{y})(x^2-xy+y^2) = 9##

i just looked at it and said the nicest reciprocals that add up to 3/4 are 1/2 and 1/4. fortunately 2^3 + 4^3 is also divisible by 9.

stevendaryl said:
That was my hint, except in the alternative form:
##(\frac{1}{x}+\frac{1}{y})(x^2-xy+y^2) = 9##
I deleted my post which was edited by the moderator, as it did not give any new hint in that form.
The equation x3+y3=9xy was written in the OP (multiplied by 9). No need to use the reciprocals.
The factorization of x3+y3 is well known, taught in the 9th class.

Last edited:
mathwonk said:
i just looked at it and said the nicest reciprocals that add up to 3/4 are 1/2 and 1/4. fortunately 2^3 + 4^3 is also divisible by 9.
It is easy to get that solution and the other one systematically, by using new variables v=x+y, z=xy.

Last edited:
ehild said:
I deleted my post which was edited by the moderator, as it did not give any new hint in that form.
The equation x3+y3=9xy was written in the OP (multiplied by 9). No need to use the reciprocals.
The factorization of x3+y3 is well known, taught in the 9th class.

The reciprocals were already in the original statement of the problem.

I got it though...i won't show all the steps
##(x^3+y^3)/(xy)=1##
##(x+y)(x^2-xy+y^2)=x^3+y^3##
and
##(x+y)/(xy)=3/4##
##(x+y)(x^2-xy+y^2)/9=4/3 (x+y)##
##x^2-xy+y^2=12##
##(9/16)z^2-3z-12=0##
##z=8, z=-8/3##
##xy=8, xy=-8/3##
##x+y=3/4##
##x+y=6##
##x=6-y##
##y=2, x=4##
##y=4, x=2##
##y=-1+(11/3)^{0.5} , x=-1-(11/3)^{0.5}##
##y=-1-(11/3)^{0.5}, x=-1+(11/3)^{0.5}##

• epenguin
chwala said:

## Homework Statement

solve ## x^2/y + y^2/x=9, 1/x+1/y=3/4##[/B]

## The Attempt at a Solution

##9x^3+9y^3=81xy##
## 4x+4y-3xy=0##
##x^3+y^3-9xy=0, 12x+12y-9xy=0##
##x^3-12x+y^3-12y=0## [/B]
yep this does not lead anywhere.....

chwala said:
I got it though...i won't show all the steps
##(x^3+y^3)/(xy)=1 ##, ##(x^3+y^3)/(xy)=9
##(x+y)(x^2-xy+y^2)=x^3+y^3##
and
##(x+y)/(xy)=3/4##
##(x+y)(x^2-xy+y^2)/9=4/3 (x+y)##
##x^2-xy+y^2=12##
##(9/16)z^2-3z-12=0##
##z=8, z=-8/3##
##xy=8, xy=-8/3##
##x+y=3/4##
##x+y=6##
##x=6-y##
##y=2, x=4##
##y=4, x=2##
##y=-1+(11/3)^{0.5} , x=-1-(11/3)^{0.5}##
##y=-1-(11/3)^{0.5}, x=-1+(11/3)^{0.5}##
There is a mistake in the first line, and you have to indicate what z is, and how did you get the equation ##(9/16)z^2-3z-12=0##.
Otherwise, it is correct.

• chwala
ehild said:
There is a mistake in the first line, and you have to indicate what z is, and how did you get the equation ##(9/16)z^2-3z-12=0##.
Otherwise, it is correct.
yeah i see that mistake..
##x^2-xy+y^2=12##
→##(x+y)^2-3xy=12##
we know that ##x+y =(3/4)xy##
you still cannot see?

chwala said:
yeah i see that mistake..
##x^2-xy+y^2=12##
→##(x+y)^2-3xy=12##
we know that ##x+y =(3/4)xy##
you still cannot see?
I see now • chwala