3
$\begingroup$

I'm doing some Cambridge STEP papers and have come across a tricky set of equations.

\begin{align*} 99 &= c^3 + 6 cd^2 \tag{1} \\ 70 &= 3c^2d + 2d^3 \tag{2} \end{align*}

From looking around, I've found the easiest way to solve these for the real solutions is to note that $ c^3 < 99 $ and $ 2d^3 < 70 $ and then to work numerically. My question is: How can these be solved algebraically ?

Many thanks!

$\endgroup$
0

4 Answers 4

5
$\begingroup$

Find $$ \left( c + d \sqrt 2 \right)^3 = 99 + 70 \sqrt 2 $$ and $$ \left( c - d \sqrt 2 \right)^3 = 99 - 70 \sqrt 2 $$

There is a Pell equation after that, as $$ 99^2 - 2 \cdot 70^2 = 1. $$ In the language of real quadratic fields, this says $c^2 - 2 d^2 = 1$ as well. Indeed, the first one with integers is $c=3, d=2,$ and that works.

If you don't like fields, you still can use $$ \left( c^2 - 2 d^2 \right)^3 = \left( c + d \sqrt 2 \right)^3 \left( c - d \sqrt 2 \right)^3 = \left(99 + 70 \sqrt 2 \right) \left(99 + 70 \sqrt 2 \right) = 1. $$

$\endgroup$
1
  • $\begingroup$ I was noticing that it looked kind of binomial, but couldn't make it work out. Nice! $\endgroup$
    – pjs36
    Dec 21, 2015 at 23:21
2
$\begingroup$

Hint...try dividing both equations by $d^3$, and then making a substitution $x=\frac cd$.

This gives $$x^3+6x=\frac{99}{d^3}$$ and $$3x^2+2=\frac{70}{d^3}$$ Eliminating the terms involving $\frac {1}{d^3}$ you ontain the polynomial $$70x^3-297x^2+420x-198=0$$ whose only real root is $x=\frac 32$ which by substitution leads to $c=3, d=2$. But I don't know if this is the best way...

$\endgroup$
2
  • $\begingroup$ Not sure if that's going to work...And you are going in to a lot of problems. Recheck your work. $\endgroup$ Dec 21, 2015 at 22:20
  • $\begingroup$ Note that by your method, we could have had $c=6,d=4$ as well. Might want to notice that. $\endgroup$ Dec 22, 2015 at 15:47
0
$\begingroup$

Start by multiplying (1) by $3d$ and multiply (2) by $c$. We get:$$297d=3c^3d+28cd^3$$$$70c=3c^3d+2cd^3$$Subtract straight down:

$$297d-70c=26cd^3$$

Solving for $c$, we get:$$297d=26cd^3+70c=(26d^3+70)c$$$$c=\frac{297d}{26d^3+70}$$

Substitute this back into (2):$$70=3(\frac{297d}{26d^3+70})^2d+2d^3$$$$70=\frac{264627d^3}{676d^6+3640d^3+4900}+2d^3$$What a mess!

Let's try multiplying by the greatest common denominator:$$47320d^6+254800d^3+34300=264627d^3+1352d^9+7280d^6+9800d^3$$With a little rearranging:$$0=1352d^9-40040d^6+19627d^3-34300$$

Make the substitution $x=d^3$.

$$0=1352x^3-40040x^2+19627x-34300$$

I do not believe there are common factors, so we will use the cubic equation (much like the quadratic equation)

$$x=\sqrt[3]{q+\sqrt{q^2+(r-p^2)^3}}+\sqrt[3]{q-\sqrt{q^2+(r-p^2)^3}}+p$$$$p=-\frac b{3a},q=p^3+\frac{bc-3ad}{6a2},r=\frac c{3a}$$$$a=1352,b=-40040,c=19627,d=-34300$$

I don't want to do this, personally. After this, substitute back in to solve for $d^3$, then $c$.

$\endgroup$
0
$\begingroup$

\begin{gather} 99=c^3+6cd^2\\ d=\sqrt{\frac{99-c^3}{6c}} \end{gather} Substituting into the other equation yields \begin{gather} 3c^2\sqrt{\frac{99-c^3}{6c}}+2\left(\frac{99-c^3}{6c}\right)^{\frac{3}{2}}=70\\ \sqrt{\frac{99-c^3}{6c}}\left(3c^2+\frac{198-2c^3}{6c}\right)=70\\ \sqrt{\frac{99-c^3}{6c}}\left(\frac{16c^3+198}{6c}\right)=70\\ \left(\frac{99-c^3}{6c}\right)\left(\frac{16c^3+198}{6c}\right)^2= 4900\\ (-c^3+99)(16c^3+198)^2-4900(6c)^3=0\\ -256c^9+19008c^6-470340c^3+3881196=0\\ \end{gather} which is a rather tedious polynomial to solve. It can be factored using the rational roots theorem, but I will not do that here. Using a calculator gives me the only real root $c=3$. Substituting back into the expression for $d$ gives $$d=\sqrt{\frac{99-27}{18}}=\sqrt{\frac{72}{18}}=\sqrt{4}=2$$ So a solution of this system of equations is $(c,d)=(3,2)$. $d=-2$ does not work because the second equation will yield $70=-70$.

$\endgroup$
1
  • $\begingroup$ Bleh. Don't you abhore such polynomials? $\endgroup$ Dec 22, 2015 at 15:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .