If $a$ and $b$ are real numbers such that $a^3-3ab^2=44$ and $b^3-3a^2b=8$ what is value of $a^2+b^2$? I have tried by adding and subtracting these equations, but can't find anything.
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Notice $$(a+ib)^3 = (a^3-3ab^2) + (3a^2b-b^3)i = 44-8i$$ Multiply both sides by their complex conjugates, one find $$(a^2+b^2)^3 = 44^2 + 8^2 = 2000\quad\implies\quad a^2 + b^2 = \sqrt[3]{2000} = 10\sqrt[3]{2}$$
Update
If one really want to hide the use of complex numbers, one can take the squares of both conditions, sum them and then employ the familiar $1,3,3,1$ pattern to factorize the result.
$$\begin{array}{rllll} (a^3 - 3ab^2)^2 &= a^6 &- 6a^4b^2 &+ 9a^2b^4 & &= 44^2\\ (b^3 - 3a^2b)^2 &= &+ 9a^4b^2 &- 6a^2b^4 &+ b^6 &= 8^2\\ \end{array}\\ \Downarrow \rlap{\color{gray}{\leftarrow\text{ sum the 2 equations }}}\\ a^6 + 3a^4b^2 + 3a^2b^4 + b^6 = 44^2+8^2 = 2000\\ \Downarrow \rlap{\color{gray}{\leftarrow\text{ use the 1, 3, 3, 1 pattern on LHS }}}\\ (a^2 + b^2)^3 = 2000\\ \Downarrow\\ (a^2 + b^2 ) = \sqrt[3]{2000} = 10\sqrt[3]{2} $$
achille hui
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1is there a way to answer the question without complex numbers? – Airdish Jan 08 '16 at 18:10
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@S.Mo - maybe some further insight on the context of this question (i.e. where did you find it??) would help us help you – The Chaz 2.0 Jan 08 '16 at 18:15
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1@TheChaz2.0 umm, it's not my question. I'm just a happy bystander who's impressed with the answer, but looking for a real number alternative. – Airdish Jan 08 '16 at 18:16
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@S.Mo - hah! My mistake. Well if you do find the OP, please direct them to my comment :) – The Chaz 2.0 Jan 08 '16 at 18:19