Question

lf the expression $a{x}^2+b{y}^2+c{z}^2+2ayz+2bzx+2cxy$ can be resolved into rational fractions then

• A. $a^3+b^3+c^3=3abc$
• B. $a+b+c=abc$
• C. $a^2+b^2+c^2=abc$
• D. $a^2+b^2+c^2=3abc$

Answer 1

The correct option is A $a^3+b^3+c^3=3abc$

$a(y+z{)}^2+b(z+x{)}^2+c(x+y{)}^2=(a+c)y^2+(b+a)z^2+(c+b)x^2+2ayz+2bzx+2cxy$
Comparing with given equations is
$a+c=b$
$b+a=c\Rightarrow a+b+c=0$
$c+b=a$
$\therefore a^3+b^3+c^3=3abc.$