Question

Find the value of $\frac{a^3+b^3+c^3-3abc}{ab+bc+ca-a^2-b^2-c^2}$, when $a=-5,b=-6,c=10$.

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $1$

Given, a = - 5, b = - 6, c = 10
Since, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$\therefore \frac{a^3+b^3+c^3-3abc}{ab+bc+ca-a^2-b^2-c^2}=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{ab+bc+ca-a^2-b^2-c^2}$
$=\frac{(a+b+c)(a^2+b^2+c^2-ab-bc-ca)}{-(a^2+b^2+c^2-ab-bc-ca)}$
$=-(a+b+c)$
$=-(-5-6+10)$
$=1$
$\therefore$ Option A is correct.