Question

If $x+y+z=0$, then the value of $x^3+y^3+z^3=$

• A. $x^3+y^3$
• B. $x^3-y^3$
  
• C. $3xyz$
• D. $x^3+z^3$

Answer 1

The correct option is C $3xyz$

Consider $\frac{1}{2}(x+y+z)\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$
$=\frac{1}{2}(x+y+z)\left[\right(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)$
$=\frac{1}{2}(x+y+z)[2x^2+2y^2+2z^2-2xy-2yz-2zx]$
$=\frac{1}{2}(x+y+z)2(x^2+y^2+z^2-xy-yz-zx]$
$=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$
$=x^3+y^3+z^3-3xyz$
$\therefore x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]$ ...(1)
Given, $x+y+z=0$
Therefore (1) becomes,
$x^3+y^3+z^3-3xyz=0$
$\Rightarrow x^3+y^3+z^3=3xyz$