Question

If the x+y+z=15,xy+yz+zx=71and xyz=10 find the value of x³+y³+z³

Answer 1

Consider,

$\small\longrightarrow x+y+z=15$

Squaring both sides,

$\small\longrightarrow(x+y+z)^2=15^2$

$\small\longrightarrow x^2+y^2+z^2+2(xy+yz+zx)=225$

Given that $\smallxy+yz+zx=71.$ Then,

$\small\longrightarrow x^2+y^2+z^2+2\times71=225$

$\small\longrightarrow x^2+y^2+z^2+142=225$

$\small\longrightarrow x^2+y^2+z^2=83$

Now consider the following formula.

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

Putting the values $\smallx+y+z=15,\ xy+yz+zx=71,\ xyz=10$ and $\smallx^2+y^2+z^2=83,$

$\small\longrightarrow x^3+y^3+z^3-3\times10=15\left(83-71\right)$

$\small\longrightarrow x^3+y^3+z^3-30=180$

$\small\text{ \longrightarrow\underline{\underline{x^3+y^3+z^3=210}} }$