MHB How to Solve This Week's POTW Quadruple Equations?

[anemone]

Here is this week's POTW:

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Find all quadruples $(a,\,b,\,c,\,d)$ of real numbers that simultaneously satisfy the following equations:

$$\begin{array}{rcr}ab^2+cd^2\hspace{-10px} & = & \hspace{-10px}-6 \\ a^2b+c^2d\hspace{-12px} & = & 0 \\ a^3+c^3\hspace{-10px} & = & 2 \\ b^3+d^3\hspace{-10px} & = & 1 \end{array}$$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to Opalg for his correct solution:), which you can find below:

Spoiler

For any real number $x$, $$\begin{aligned} (ax+b)^3 + (cx+d)^3 &= (a^3 + c^3)x^3 + 3(a^2b + c^2d)x^2 + 3(ab^2 + cd^2)x + (b^2+d^2) \\ &= 2x^3 -18x + 1. \end{aligned}$$ The cubic polynomial $p(x) = 2x^3 - 18x + 1$ has three distinct real roots (close to, but not equal to, $-3$, $0$ and $3$). But if $(ax+b)^3 + (cx+d)^3 = 0$ then $(ax+b)^3 = -(cx+d)^3$. Taking cube roots, $(ax+b) = -(cx+d)$, so that $x(a+c) + (b+d) = 0$. The only way for that linear equation to hold for three distinct values of $x$ is if $a+c=0$. But then $a^3+c^3 = 0$, contradicting one of the original equations.

Conclusion: there are no real solutions $(a,b,c,d)$ to the given equations.