Find all possible values of $a^3 + b^3$ if $a^2+b^2=ab=4$.
From $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(4-4)=(a+b)0$. Then we know $a^3+b^3=0$. If $a=b=0$, it is conflict with $a^2+b^2=ab=4$. If $a\neq0$ and $b\neq0$, then $a$ and $b$ should be one positive and one negative. This contradict with $ab$=4.
I don't know the solution.
Any help please.
There are no solutions in the real numbers.
Plotting $a^2+b^2=4$ on a graph, we get a circle radius 2. Plotting $ab=4$ on the same graph, we get a hyperbola which goes through $(2,2)$ (with the negative side going through $(-2,-2)$) which does not intersect the circle.
A tag of "recreational mathematics" wouldn't usually cause me to go looking for complex solutions unless they were apparent from the setup. However Lythia's answer correctly identifies that the options in the complex plane are restricted to $a$ and $b$ choosing from the $6^{th}$ roots of $-64$. Since $ab=4$, $a$ and $b$ must be complex conjugates. The $a^2+b^2=4$ requirement additionally means that $a$ and $b$ cannot be $\pm 2i$, giving the four answer pairs of $a=\pm \sqrt 3\pm i$ and $b=a^*$.
Allowing complex $a$ and $b$ rescues the question's original analysis and reinstates the answer of $a^3+b^3 = 0$.
To give a different perspective, I can't find anywhere in the problem statement which states $a$ and $b$ must be real. Therefore, assume $a,b \in \mathbb{C}$. Then the equation $a^3+b^3=0$ along with the condition $ab=4$ implies, by multiplying both sides by either $a^3$ or $b^3$ that $a^6+64 = 0$ and $b^6+64=0$. Therefore, $$ a = \pm\sqrt[6]{-64} = \pm2\sqrt[6]{-1},\\ b= \pm\sqrt[6]{-64} = \pm2\sqrt[6]{-1}.$$ As $a$ and $b$ are multiples of sixth roots of unity, there are six choices for each. Not all choices are valid since the condition $a^2+b^2=4$ must also be met. For example $$a=b=\sqrt{3}+i$$ does not work since then $a^2+b^2=4+4\sqrt{3}i$. However, the choice $$a=\sqrt{3}+i, b=\sqrt{3}-i$$ works fine.
Suppose there were real solutions to $a^2 + b^2 = ab = 4$. Then $$0 \leq (a - b)^2 = a^2 - 2ab + b^2 = (a^2 + b^2) - 2ab = -4$$ which is a a contradiction. Thus there cannot be real $a, b$ satisfying the requirements.
It is given that $a^{2} + b^{2} = ab =4$ and the goal is to determine all values of $a^{3} + b^{3}$. Now, it is seen that $a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2}) = 0$. So far the set of equations becomes \begin{align} a^{3} + b^{3} &= 0 \\ a^{2} + b^{2} &= 4 \\ ab &= 4. \end{align} From the first, multiply by $a^{3}$ to obtain $0 = a^{6} + (ab)^{3} = a^{6} + 2^{6}$ which yields $a = \pm 2 e^{\pi i/6 \pm 2n \pi i}$. Since $ab=4$ then yields $b = \pm 2 e^{-\pi i/6 \pm 2n\pi i}$. In both cases $n \geq 0$. The possible values of $a$ are \begin{align} a \in \{ 2 e^{\pi i/6 + 2n \pi i}, 2 e^{\pi i/6 - 2n \pi i}, - 2 e^{\pi i/6 + 2n \pi i}, - 2 e^{\pi i/6 - 2n \pi i} \} \end{align} and for $b$ \begin{align} b \in \{ 2 e^{-\pi i/6 + 2n \pi i}, 2 e^{-\pi i/6 - 2n \pi i}, - 2 e^{-\pi i/6 + 2n \pi i}, - 2 e^{-\pi i/6 - 2n \pi i} \} \end{align} In view of these two sets of values it is seen that the sets must be taken in order of signs, ie the sets $(a, b) \in \{(+,+), (-,-)\}$, \begin{align} a^{3} + b^{3} &= 2 (\pm 2)^{3} \cos\left(\frac{\pi}{2}\right) = 0 \\ a^{2} + b^{2} &= 2 (\pm 2)^{2} \cos\left(\frac{\pi}{3}\right) = 4 \\ ab &= (\pm 2)^{2} = 4. \end{align}
As has been observed, one approach to this problem is to solve the equations over the complex numbers, and then check directly that $a^3+b^3=0$ in all cases. However, it's both useful and more general to observe that the conclusion can be reached without actually solving the equations: since $a^2+b^2=a b$, we have
$(*)\ \ a^3+b^3=(a+b)(a^2-a b + b^2)=0.$
The problem makes sense over any field -- in fact in any commutative ring with $1$ (this condition is only needed so that "$4$" makes sense.) In addition to being simple, an advantage of the solution $(*)$ is that it works regardless of the setting. This is relevant, since the problem did not specify a domain for $a$ and $b$, so a solution based on calculating the roots in a particular context involves an inherent assumption.
For example, in the field ${\Bbb F}_{37}$ (i.e., the integers mod $37$), the equations have the four solutions $(a,b)$=$(9,21)$, $(16,28)$, $(21,9)$, and $(28,16)$. You can check that $a^3+b^3=0$ in each case, but you're back to square one when faced with ${\Bbb F}_{61}$.
