Question

if X + Y + Z = 6 and X square + Y square + Z square = 18 then find x cube + y cube + Z cube - 3 x y z​

Answer 1

$\textbf{Given:}$

$x+y+z=6$

$x^2+y^2+z^2=18$

$\text{We know that}$

$\boxed{\bf\;(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)}$

$\implies(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)}$

$\implies(6)^2=18+2(xy+yz+zx)$

$\implies\;36-18=2(xy+yz+zx)$

$\implies\;18=2(xy+yz+zx)$

$\implies\;xy+yz+zx=9$

$\text{We know that}$

$\boxed{\bf\;a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))}$

$\implies\;x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))$

$\implies\;x^3+y^3+z^3-3xyz=(6)(18-9)$

$\implies\;x^3+y^3+z^3-3xyz=(6)(9)$

$\implies\boxed{\bf\;x^3+y^3+z^3-3xyz=54}$