Question

If $x=a^2-bc,y=b^2-ca,z=c^2+ab$, then value of $\frac{(a+b+c)(x+y+z)}{ax+by+cz}$ is equal to:

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (C)

$1$

$(a+b+c)(x+y+z)$

$=(a+b+c)[a^2-bc+b^2-ca+c^2-ab]$

$=(a+b+c)[a^2+b^2+c^2-bc-ca-ab]$
$=a^3+b^3+c^3-3abc$ ...(i).

Now, $ax+by+cz=a(a^2-bc)+b(b^2-ca)+c(c^2-ab)$
$=a^3+b^3+c^3-3abc$ ...(ii).

$\therefore \frac{\left(i\right)}{\left(ii\right)}=1$.

The value of the given expression is $1$.

Therefore, option $C$ is correct.