Cho \(\frac{{{x^2} - yz}}{a} = \frac{{{y^2} - zx}}{b} = \frac{{{z^2} - xy}}{c}\). Chứng minh rằng \(\frac{{{a^2} - bc}}{x} = \frac{{{b^2} - ca}}{y} = \frac{{{c^2} - ab}}{z}\).

Hướng dẫn giải

Ta có: \(\frac{{{x^2} - yz}}{a} = \frac{{{y^2} - zx}}{b} = \frac{{{z^2} - xy}}{c}\)

Suy ra \(\frac{a}{{{x^2} - yz}} = \frac{b}{{{y^2} - zx}} = \frac{c}{{{z^2} - xy}}\)

Suy ra \({\left( {\frac{a}{{{x^2} - yz}}} \right)^2} = {\left( {\frac{b}{{{y^2} - zx}}} \right)^2} = {\left( {\frac{c}{{{z^2} - xy}}} \right)^2}\)

Lại có \[\frac{{{a^2}}}{{{{\left( {{x^2} - yz} \right)}^2}}} = \frac{{bc}}{{\left( {{y^2} - zx} \right)\left( {{z^2} - xy} \right)}} = \frac{{{a^2} - bc}}{{\left( {{x^4} - 2{x^2}yz + {y^2}{z^2}} \right) - \left( {{y^2}{z^2} - x{y^3} - x{z^3} + {x^2}yz} \right)}}\]

\[ = \frac{{{a^2} - bc}}{{{x^4} - 3{x^2}yz + x{y^3} + x{z^3}}} = \frac{{{a^2} - bc}}{{x\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}}\]

Tương tự \[\frac{{{b^2}}}{{{{\left( {{y^2} - zx} \right)}^2}}} = \frac{{{b^2} - ac}}{{y\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}}\];

\[\frac{{{c^2}}}{{{{\left( {{z^2} - xy} \right)}^2}}} = \frac{{{c^2} - ab}}{{z\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}}\]

Suy ra \[\frac{{{a^2} - bc}}{{x\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}} = \frac{{{b^2} - ac}}{{y\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}} = \frac{{{c^2} - ab}}{{z\left( {{x^3} + {y^3} + {z^3} - 3xyz} \right)}}\]

Do đó \[\frac{{{a^2} - bc}}{x} = \frac{{{b^2} - ac}}{y} = \frac{{{c^2} - ab}}{z}\].