Question

(a) If $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ then prove that
any two of a,b,c must be equal and opposite. All the quantities a,b,c being assumed to be real.
(b) Resolve into factors the expression
${(a+b+c)}^3-a^3{-b}^3-c^3$

Answer 1

(a) The given relation can be written as
$(a+b+c)(ab+bc+ca)=abc$
or $a^2+(b+c)+abc+b^2(c+a)+abc+c^2(a+b)+abc=abc$
or $a^2+(b+c)+b^2(c+a)+c^2(a+b)+2abc=0$
Above consists of $8$ terms taking
$2abc=abc+abc$
L.H.S. is $(a+b)(b+c)(c+a)=0$
It is satisfied if either $a+b=0$ or $b+c=0$ or $c+a=0$ i.e. any two of the three quantities are equal and opposite.
(b) The given expression becomes 0 when either $a+b=0$ or $b+c=0$ or $c+a=0$ and hence all these are its factors of 3rd degree.
Now assume
${(a+b+c)}^3-a^3-b^3-c^3=\lambda (a+b)(b+c)(c+a)$
Choose $a=0,b=1,c=2$, we get $\lambda =3$.