I am struggling to solve these simultaneous equations:
a²+ab+b²=7
b²+bc+c²=28
c²+ac+a²=21
Any help would be highly appreciated.
The first thing I would do is to test if any of the unknowns are 0 or 1 because that would make life much easier.
If that does not work, I'd move to the quadratic formula to isolate the unknowns so that I can make substitutions.
\(\displaystyle a^2 + ab + b^2 = 7 \implies a^2 + ab + b^2 - 7 = 0 \implies\)
\(\displaystyle a = \dfrac{-\ b \pm \sqrt{b^2 - 4(1)(b^2 - 7)}}{2} = \dfrac{-\ b \pm \sqrt{28 - 3b^2}}{2}.\)
\(\displaystyle \text {Similarly, } b^2 + bc + c^2 = 28 \implies c = \dfrac{-\ b \pm \sqrt{112 - 3b^2}}{2}.\)
We can now attack the third equation as an equation in one unknown.
\(\displaystyle c^2 + ac + a^2 = 21 \implies\)
\(\displaystyle \left ( \dfrac{-\ b \pm \sqrt{112 - 3b^2}}{2} \right )^2 + \dfrac{-\ b \pm \sqrt{28 - 3b^2}}{2} * \dfrac{-\ b \pm \sqrt{112 - 3b^2}}{2} + \left ( \dfrac{-\ b \pm \sqrt{28 - 3b^2}}{2} \right )^2 = 21.\)
Solving the four equations implied by that mess may be arduous, but it should be possible because they will turn out to be quartics and those are in principle solvable. (If, that is, there are are any real solutions).
EDIT: If you know that the answers are supposed to integers, you can tell that the absolute value of b is small because
\(\displaystyle 3b^2 \le 28 \implies b= -\ 3, -\ 2, -\ 1, 0, 1,\ 2 \text { or } 3.\)
Then you can use trial and error. For example,
\(\displaystyle b = 2 \implies a^2 + 2a + 4 = 7 \implies a^2 + 2a - 3 = 0 \implies\)
\(\displaystyle (a + 3)(a - 1) = 0 \implies a = -\ 3 \text { or } a = 1.\)
\(\displaystyle a = 1 \implies a^2 + ac + c^2 = 21 \implies 1 + c + c^2 = 21 \implies\)
\(\displaystyle c^2 + c - 20 = 0 \implies (c + 5)(c - 4) \implies c = -\ 5 \text { or } c = 4.\)
\(\displaystyle b^2 + bc + c^2 = 4 + 8 + 16 = 28.\)