Prove that $\sqrt{18} - \sqrt{12}-\sqrt{45} + \sqrt{6}$ is irrational

Question

Prove that $\sqrt{18} - \sqrt{12}-\sqrt{45} + \sqrt{6}$ is irrational

I tried to let $x = \sqrt{18} - \sqrt{12}-\sqrt{45} + \sqrt{6}$, and assume for the sake of contradiction that $x$ is rational.
Then I squared numerous times to attempt to isolate an irrational on one side, and a supposedly rational expression on another side, since we assumed $x$ is rational, to arrive at a contradiction.
However, all of my attempts have resulted in very complicated expressions that I am unable to complete.

Is this the right approach, or are there other methods such as looking at it as a polynomial root?

Answer 1

To assist OP & at the risk of downvoting , I will add a very simple answer which the Particular numbers involved here allow (The Duplicates do not seem to allow this method) . . . .

Let $X=\sqrt{18} - \sqrt{12}-\sqrt{45} + \sqrt{6}=3\sqrt{2} - 2\sqrt{3}-3\sqrt{5} + \sqrt{2}\sqrt{3}$

Hence $X - 3\sqrt{2} + 2\sqrt{3} - \sqrt{2}\sqrt{3} = - 3\sqrt{5}$
Squaring will eliminate $\sqrt{5}$ , while retaining $\sqrt{2}$ & $\sqrt{3}$ & $\sqrt{2}\sqrt{3}$

Move the terms around to get :
$\square 1+\square \sqrt{3} = \square \sqrt{2} + \square \sqrt{2}\sqrt{3} = \sqrt{2} ( \square + \square \sqrt{3} )$ , where $\square$ indicates some arbitrary Co-Efficients involving $X$ , $X^2$ & Constants , though not involving radicals.

Square this to eliminate $\sqrt{2}$ , while retaining $\sqrt{3}$ on either side.

Collect all $\sqrt{3}$ terms to make it $\sqrt{3}=\square/\square$ which is a Contradiction.
[[ Checking that it is not $0/0$ is left out here , though it is not very hard ]]

Summary / Overview :

This is a Proof By Contradiction.
Assume that the given $X$ is rational.
What will that tell us about $X^2$ ? That too must be rational.
What about $X^2-6X+8$ ? Rational too. What about $1/(X^2-6X+8)$ ? Rational too.
Basically , we can Add , Subtract , Divide , Multiply (& Square !) rational numbers to get new rational numbers.
What occur when we have $Y=(X^2+12X+32)/(X^2+3X+2)$ ?
That $Y$ must be rational too.
Well , when we have $\sqrt{3}=(X^2+12X+32)/(X^2+3X+2)$ , we will get Contradiction that $\sqrt{3}$ is rational , which is not true.

What will that Contradiction tell us about our original assumption ?
Well , that assumption must be not true : We then know that the given $X$ is not rational.

Answer 2

HINT.-You have to prove that $$3\sqrt2-2\sqrt3-3\sqrt5+\sqrt6$$ Show first that $x=\sqrt a+\sqrt b$ is irrational for $a,b$ both integers square-free. It is very known that $x$ is of degree $4$ over $\Bbb Q$ so $x$ is irrational (its minimal polynomial is $x^4-2(a+b)x^2+(a-b)^2=0$.

It follows that, for example, $x=3\sqrt2-2\sqrt3$ and $y=-3\sqrt5+\sqrt6$ are both irrational so its sum, $x+y$, is rational if and only if the sum of its fractional parts is integer. You can verify that this is not so with a little calculation of only two decimal digits.