# Given system of equations $a+b = 2, ab=4$ solve $a^2+b^2=?$ and $a^3=?$

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.

• $(a+b)^2=a^2+b^2+2ab$ so $(a+b)^2-2ab=a^2+b^2$ and you know the value of both $(a+b)^2$ and $ab$, geez i wonder what could you do... Jul 21, 2014 at 12:47
• x^2-(a+b)x+ab=0 has the roots a and b. Jul 21, 2014 at 13:32

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.

• in your second hint, how do you arrive to $a^3$? Jul 21, 2014 at 12:51
• After you get the sum, which I think will be -16, you need to back substitute b=4/a and solve. :)
– MonK
Jul 21, 2014 at 12:55
• the sum is -16 i think Jul 21, 2014 at 13:02
• Yeah! sorry about that, got mixed up with my work. it is -16
– MonK
Jul 21, 2014 at 13:02

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.

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 $-8 = (1+\sqrt{3}i)^3 = (1-\sqrt{3}i)^3$