This is the following question:
Suppose that $a, b, c$ are integers such that
$a\sqrt{2} + b = c\sqrt{3}$
(i) By squaring both sides of the equation, show that $a = b = c = 0$
The answer says that you put the equation into the following form:
if $ab \neq 0$
$\sqrt{2} = \frac{3c^2 − 2a^2 − b^2}{2ab}$
is rational — a contradiction and so a = 0 or b = 0.
Why would $a$ or $b$ be 0? (I get that you cannot express an irrational number as the quotient of two rational numbers).
If $a,b \in \mathbb{Z}$, then $$ 2a^{2} + b^{2} + 2ab\sqrt{2} = 3c^{2}, $$ and then \begin{align} (*)\ \ \ \ ab\sqrt{2} = \frac{3c^{2}-2a^{2}-b^{2}}{2} . \end{align} The number $\sqrt{2}$ is irrational; so $ab \neq 0$ leads to a contradiction, and hence $ab = 0$. We claim that $a=b=c = 0$; without loss of generality, let $a = 0$ and let $b \neq 0$. Then from $(*)$ we have $|b| = \sqrt{3}|c|$, which, by the fact that $\sqrt{3}$ is irrational, shows that $bc \neq 0$ is absurd; hence $bc = 0$. But either $b \neq 0$ or $c \neq 0$ also gives absurdity, so $b=c=0$.