<s>[INST] Value of $a^2 + b^2$ given $a^3 - 3ab^2 = 44$ and $b^3 - 3a^2 b = 8$ 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. 
 [/INST] 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.
$$\begin{array}{rllll}
(a^3 - 3ab^2)^2 &= a^6 &- 6a^4b^2 &+ 9a^2b^4 & &= 44^2\\
(b^3 - 3a^2b)^2 &=     &+ 9a^4b^2 &- 6a^2b^4 &+ b^6 &= 8^2\\
\end{array}\\
\Downarrow \rlap{\color{gray}{\leftarrow\text{ sum the 2 equations }}}\\
a^6 + 3a^4b^2 + 3a^2b^4 + b^6 = 44^2+8^2 = 2000\\
\Downarrow \rlap{\color{gray}{\leftarrow\text{ use the 1, 3, 3, 1 pattern on LHS }}}\\
(a^2 + b^2)^3 = 2000\\
\Downarrow\\
(a^2 + b^2 ) = \sqrt[3]{2000} = 10\sqrt[3]{2}
$$
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