Question

Cho :

x+y+z=a

x²+y²+z²=b²

^{\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\)}

Tính x³+y³+z³ theo a,b,c

Answer 1

Ta có: \(\left(x+y+z\right)=a\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=a^2\)

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

Ta lại có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}\)

\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{c}\)

\(\Leftrightarrow xyz=c\left(xy+yz+zx\right)=c.\frac{a^2-b^2}{2}\)

Ta biến đổi: \(x^3+y^3+z^3=x^3+y^3+z^3-3xyz+3xyz\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-\left(xy+yz+zx\right)\right)+3xyz\)

\(=a.\left(b^2-\frac{a^2-b^2}{2}\right)+\frac{3c\left(a^2-b^2\right)}{2}\)