Question Number 184607 by mr W last updated on 09/Jan/23
$${solve} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{16}}\\{{y}^{\mathrm{2}} −{yz}+{z}^{\mathrm{2}} =\mathrm{25}}\\{{z}^{\mathrm{2}} −{zx}+{x}^{\mathrm{2}} =\mathrm{49}}\end{cases} \\ $$
Commented by Frix last updated on 09/Jan/23
$$\mathrm{Exact}\:\mathrm{solution}\:\mathrm{possible}\:\mathrm{but}\:\mathrm{not}\:\mathrm{useful}… \\ $$
Commented by MJS_new last updated on 11/Jan/23
$$\mathrm{the}\:\mathrm{system}\:\begin{cases}{{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} ={a}^{\mathrm{2}} }\\{{y}^{\mathrm{2}} +{yz}+{z}^{\mathrm{2}} ={b}^{\mathrm{2}} }\\{{z}^{\mathrm{2}} +{zx}+{x}^{\mathrm{2}} ={c}^{\mathrm{2}} }\end{cases}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$$$\mathrm{like}\:\mathrm{this}: \\ $$$$\mathrm{1}.\:\mathrm{solve}\:\left(\mathrm{2}\right)−\left(\mathrm{3}\right)\:\mathrm{for}\:{z} \\ $$$$\mathrm{2}.\:\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{3}.\:\mathrm{let}\:{y}={px}\:\mathrm{and}\:\mathrm{solve}\:\left(\mathrm{1}\right)\:\mathrm{for}\:{x}^{\mathrm{2}} \\ $$$$\mathrm{4}.\:\mathrm{insert}\:\mathrm{in}\:\left(\mathrm{2}\right)\:\mathrm{which}\:\mathrm{leads}\:\mathrm{to}\:{p}^{\mathrm{2}} +\alpha{p}+\beta=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{system}\:\begin{cases}{{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} ={a}^{\mathrm{2}} }\\{{y}^{\mathrm{2}} −{yz}+{z}^{\mathrm{2}} ={b}^{\mathrm{2}} }\\{{z}^{\mathrm{2}} −{zx}+{x}^{\mathrm{2}} ={c}^{\mathrm{2}} }\end{cases}\:\mathrm{leads}\:\mathrm{to} \\ $$$${p}^{\mathrm{4}} +\alpha{p}^{\mathrm{3}} +\beta{p}^{\mathrm{2}} +\gamma{p}+\delta=\mathrm{0}\:\mathrm{which}\:\mathrm{can}\:\mathrm{only}\:\mathrm{be} \\ $$$$\mathrm{exactly}\:\mathrm{solved}\:\mathrm{in}\:\mathrm{some}\:\mathrm{cases}. \\ $$$$\mathrm{with}\:{a}=\mathrm{4},\:{b}=\mathrm{5},\:{c}=\mathrm{7}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{possible}. \\ $$
Commented by mr W last updated on 11/Jan/23
$${thanks}\:{sir}! \\ $$