How do you rationalize the denominator and simplify `(3+sqrt18) /( 1 + sqrt8 )`?

A very useful relationship to remember is `a^2-b^2=(a-b)(a+b)`
In the context of this question we can change
`1+sqrt(8)" to "1^2-(sqrt(8))^2`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Multiply by 1 but in the form of `1= (1-sqrt(8))/(1-sqrt(8))`

`" "color(red)("Error - should be "sqrt(18))`
`" "color(red)(darr)`
`cancel(((3+color(red)(sqrt(8)))(1-sqrt(8)))/((1+sqrt(8))(1-sqrt(8))))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(green)("Corrected!!!")`

`((3+sqrt(18))(1-sqrt(8)))/((1+sqrt(8))(1-sqrt(8))) = (3-3sqrt(8)+sqrt(18)-sqrt(8xx18))/(1-8)`

`=(3-3sqrt(2xx2^2)+ sqrt(2xx3^2)- sqrt(2xx2^2xx2xx3^2))/(-7)`

`=(3-6sqrt(2)+3sqrt(2)-12)/(-7)`

`=(-9-3sqrt(2))/(-7)" "=" "+3/7(3+sqrt(2))`

Got there in the end! Amazing how much a typo can mess things up!

Rationalisation factor of `a+sqrtb` is `a-sqrtb` and vice-versa.

Given Exp. `=(3+sqrt18)/(1+sqrt8)=(3+sqrt(9*2))/(1+sqrt(4*2))`

`=(3+3sqrt2)/(1+2sqrt2)`

Multiplying, in `Nr. & Dr` by the rationalisation factor of Dr., i.e., by`(1-2sqrt2)`, the Exp.

`={(3+3sqrt2)(1-2sqrt2)}/{(1+2sqrt2)(1-2sqrt2)},`

`={3+3sqrt2-6sqrt2-6*2}/{1^2-(2sqrt2)^2},`

`=(-9-3sqrt2)/(1-8),=3/7(3+sqrt2).`