shmm wrote: Solve : $y(x+y)^2=9 $ ; $y(x^3-y^3)=7$ Divide $y(x+y)^2=9 $ by$y(x^3-y^3)=7$ and then expand $x^3-y^3$, cancel out the common factors. The rest is trivial

Can you post your solution??

+1 I 'm curious too

shmm wrote: Solve : $y(x+y)^2=9 $ ; $y(x^3-y^3)=7$ System implies $x>y>0$ Let $y=t^2$ with $t>0$ and first equation gives $x=\frac 3t-t^2$ Plugging this in second equation, we get $h(t)=2t^9-9t^6+27t^3+7t-27=0$ whose derivative is $h'(t)=18t^8-54t^5+81t^2+7$ $=18(t^4-\frac 32t)^2+\frac{81}2t^2+7$ $>0$ So $h(t)$ is increasing and since $h(1)=0$, $1$ is the only root of $h(t)$ Hence the solution $\boxed{(x,y)=(2,1)}$

Hi pco, excellent solution. I have a few questions on how you thought of certain steps if you don't mind pco wrote: Let $y=t^2$ with $t>0$ pco wrote: $h'(t)=18t^8-54t^5+81t^2+7$ $=18(t^4-\frac 32t)^2+\frac{81}2t^2+7$ $>0$

BigInteger wrote: Hi pco, excellent solution. I have a few questions on how you thought of certain steps if you don't mind pco wrote: Let $y=t^2$ with $t>0$ I tried to get one variable depending on the other from one equation and then to plug in the other. The best equation for that is clearly the first and we get $x+y=\frac 3{\sqrt y}$ Writing then $y=t^2$ is just a simplification for writing equation : $x=\frac 3t-t^2$ BigInteger wrote: pco wrote: $h'(t)=18t^8-54t^5+81t^2+7$ $=18(t^4-\frac 32t)^2+\frac{81}2t^2+7$ $>0$ Proving (or trying to prove) that an even-degree polynomial has no real roots is very often a matter of transforming it in a sum of squares. Considering $h'(t)=18t^8-54t^5+81t^2+7$ : One of the squared polynomials must be with degree $4$ We cant have degree $3$ summands in it (since we have no degree $7$ elements in h'(t). We can have a degree $2$ summand but this will generate the need of a cubic polynomial squared to cancel the generated degree $6$ element. So it would be simpler to try first without any degree $2$ summand. A degree $1$ summand will generate a degree $5$ element in the squared form and we need this element to be exactly the existing degree $5$ (else we need another polynomial whose square cancel it, which is impossible since $5$ is even, except chosing a cubic or above, but then we'd have problems again with degrees $6$ and above). Since coefficient of $x^5$ in $18(x^4+ax+b)^2$ is $36a$, we need $36a=-54$ and so $a=-\frac 32$ And so it remains to write : $18t^8-54t^5+81t^2+7$ $=18(t^4-\frac 32t+b)^2 -36bt^4+\frac{81}2t^2+54bt+7-18b^2$ And we must choose $b$ in such a way $-36bt^4+\frac{81}2t^2+54bt+7-18b^2$ has no real root. Choosing $b=0$ in order to have just a quadratic seems a very easy simple first choice which indeed works fine. And so $18t^8-54t^5+81t^2+7$ $=18(t^4-\frac 32t)^2+\frac{81}2t^2+7$

$(x+y)^2 y=9.1$ $\implies (x+y)^2=3^2$, $y=1$ $\implies (x+y)=3$ , (–3not satisfies) $y=1$ $(x, y) =(2,1)$ $x+y=1,y=9$ Does not satisfy the condition $y(x^3 –y^3)=1(2^3–1^3)=7$

Why on earth is this here?

OlympusHero wrote: Why on earth is this here?

And why isn't there any other solutions?