if postive real number $x,y$ such $$\begin{cases} 3x^3+4y^3=7\\ 4x^4+3y^4=16 \end{cases}$$
Find $x+y=?$
My try: $$4x^4-3x^3+3y^4-4y^3=9$$ $$x^3(4x-3)+y^3(3y-4)=9$$
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What is the source of this question? – Sawarnik Jan 17 '14 at 12:15
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This is Mathematical olympiad problem in china (2006) – Jan 17 '14 at 12:16
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Probable typo: the title has a $17$, not a $7$. – Barry Cipra Jan 17 '14 at 12:27
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1According to WA, there are no solutions where both $x$ and $y$ are positive. – Daniel R Jan 17 '14 at 12:30
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1@DanielR But the question was to find $x + y$. – Yiyuan Lee Jan 17 '14 at 12:32
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@LeeYiyuan "if postive real number x,y such..." – Daniel R Jan 17 '14 at 12:33
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No real solutions for 17 and 16... – Umberto Jan 17 '14 at 12:33
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@DanielR Woops! Missed that out. – Yiyuan Lee Jan 17 '14 at 12:34
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We can find $x^3(4x-3)$, $y^3(3y-4)$ individually but I do not see how that leads to $x+y$. – Sawarnik Jan 17 '14 at 12:35
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@nanchangjian x+y has to be a number or can depend from the variable x and y? otherwise is x+y= $\frac{16+3x^3y+4xy^3-x^4+y^4}{7}$ (i can show the passages if it can depend from x & y =)) – Bman72 Jan 17 '14 at 12:48
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2to @nanchangjian: are the numbers in the original questions correct? Os is there a typo? – Umberto Jan 17 '14 at 13:00