I am trying to solve $a^2+b^2$ and $a^3$ given $a+b = 2, ab=4$. I have the key with the answers $a^2+b^2=-4$ and $a^3=-8$ but am wondering which steps to take to get to that answer.
My understanding of the problem so far is that from this system of equations it is not possible to get integer solutions for $a$ and $b$ (which is not asked for). I have not really gotten further in solving it than this realization.
$\endgroup$ 23 Answers
$\begingroup$HINT:
$a^2+b^2=(a+b)^2-2(ab)$
and
$a^3+b^3=(a+b)^3-3ab(a+b)$
EDIT: To further elongate the second equation
$a^3+b^3=-16$
Now, substitute b=4/a and solve.
$\endgroup$ 4 $\begingroup$Hint: $$ (a+b)^2 = a^2 + 2ab + b^2 $$
Hint:
Two equations in two unknowns can be enough to determine two unknowns. Pick an equation and try to express one unknown by another and use the other equation.
$\endgroup$ $\begingroup$if all else fails, remember that you can solve for $a$ and $b$ explicitly. I get roots $1+\sqrt{3}i$ and $1-\sqrt{3}i$, and $-8 = (1+\sqrt{3}i)^3 = (1-\sqrt{3}i)^3$
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