Problem source

The first question on an examination paper is:
Solve for $x$ the equation $\displaystyle \frac 1x = \frac 1 a + \frac 1b \;.$
where (in the question) $a$ and $b$ are given non-zero real numbers.
One candidate writes $x=a+b$ as the solution. Show that there are no values of $a$ and $b$ for which this will give the correct answer.
The  next question on the  examination paper is:
Solve for $x$ the equation $\displaystyle \frac 1x = \frac 1 a + \frac 1b  +\frac 1c \;.$
where (in the question) $a\,$, $b$ and $c$  are given non-zero numbers.
The  candidate uses the same technique, giving the answer as $\displaystyle x = a + b  +c \;.$
Show that the candidate's answer will be correct if and only if  $a\,$, $b$ and $c$ satisfy at least one of the equations
$a+b=0\,$, $b+c=0$ or $c+a=0\,$.

Solution source

Suppose \begin{align*}
&& \frac{1}{a+b} &= \frac{1}{a} + \frac{1}{b} \\
\Rightarrow && ab &= b(a+b)+a(a+b) \\
&&&= (a+b)^2 \\
\Rightarrow && 0 &= a^2+ab + b^2 \\
&&&= \tfrac12 (a^2+(a+b)^2+b^2)
\end{align*}
Which clearly has no solution for non-zero $a,b$.

Suppose \begin{align*}
&& \frac{1}{a+b+c} &= \frac1a + \frac1b+\frac1c \\
\Leftrightarrow && abc &= (a+b+c)(bc+ca+ab) \\
\Leftrightarrow && 0 &= (a+b+c)(bc+ca+ab) - abc \\
&&&= (a+b)(b+c)(c+a)
\end{align*}

Therefore it is true iff $a+b = 0$ or $b+c=0$ or $c+a =0$ as required.