Question

The value of $\frac{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}{9(x-y)(y-z)(z-x)}$ is equal to__________

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

The correct option is C $\frac{1}{3}$

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2+ab+bc+ac)$

And in the above question $a+b+c=x-y+y-z+z-x=0$

So, $a^3+b^3+c^3=3abc$

$\frac{{(x-y)}^3+{(y-z)}^3+{(z-x)}^3}{9(x-y)(y-z)(z-x)}$$=\frac{3(x-y)(y-z)(z-x)}{9(x-y)(y-z)(z-x)}$

$=\frac{1}{3}$