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Let $p^3+q^3=4$ and $pq=\frac{2}{3}$ . Find $p+q$.

A graphing calculator can find values of $p$ and $q$ numerically. As one can see from the graph below, the two solutions are approximately $(0.4, 1.6)$ and $(1.6, 0.4)$:

plot of equations

However, I am interested in a symbolic solution. Is there a method to solve this problem quickly without having to use a graphing calculator?

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    $\begingroup$ You might find this useful: $p^3 + q^3 = (p+q)^3 - 3pq(p+q)$ $\endgroup$ – Tim Jan 27 '15 at 1:34
  • $\begingroup$ In general, I don't think so, as it involves the computations used for solving third degree equations by Cardano's method. $\endgroup$ – Bernard Jan 27 '15 at 1:34
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Hint:

$$p^3+q^3=4$$ $$p^3q^3=(\frac{2}{3})^3 \Rightarrow q^3=\frac{8}{27p^3}$$

The equation $$p^3+\frac{8}{27p^3}=4$$ is quadratic in $p^3$.

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$$(p+q)^3=[p^3+q^3]+3pq(p+q)$$

$$(p+q)^3=[4]+3\frac{2}{3}(p+q)$$

$$(p+q)^3=[4]+2(p+q)$$

$$u^3-2u-4=0$$ where $u=p+q$

$$(u-2)(u^2+2u+2)=0$$

Therefore $u=2$, or some complex solutions.

Note that the root 2 could have been found by the rational root theorem.

The long way: $$pq=\frac{2}{3}\implies q=\frac{2}{3p}$$ $$p^3+q^3=4$$ $$\implies p^3+\frac{8}{27p^3}=4$$ $$\implies p^6-4p^3+\frac{8}{27}=0$$ Let $u=p^3$ $$\implies u^2-4u+\frac{8}{27}=0$$ $$\implies u= 2\pm\frac{10\sqrt3}{9}$$ $$\implies p=\sqrt[3]{2\pm\frac{10\sqrt3}{9}}$$ $$\implies q=\frac{2}{3[\sqrt[3]{2\pm\frac{10\sqrt3}{9}}]}$$ $$\cdots$$ Can you take it from there?

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we have $p^3 + q^3 = 4$ and $pq = 2/3$ so $p^3 q^3 = 8/27.$ therefore $$0= (x - p^3)(x-q^3) = x^2 - (p^3 + q^3)x + p^3q^3 = x^2 - 4x + 8/27$$ solving this quadratic equation we find $x = \dfrac{4 \pm \sqrt{16 - 4*8/27}}{2} =\dfrac{6\sqrt 3 \pm 10}{ 3\sqrt 3}$

so $$p+q = \left( \dfrac{6\sqrt 3 + 10}{ 3\sqrt 3}\right)^{1/3} + \left( \dfrac{6\sqrt 3 - 10}{ 3\sqrt 3}\right)^{1/3} $$

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