<s>[INST] Rationalizing radicals Rationalize
(i.e., get rid of radicals in the denominator)
in
$\dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$,
$\dfrac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}$,
and,
in general
$\dfrac{1}{\sum_{i=1}^n \sqrt{a_i}}$.
I have submitted my work so far as an answer.
 [/INST] As you know, the identity
$$(a+b)(a-b) \;\; = \;\; a^2 - b^2$$
shows that if you start with $a+b$ and multiply by $a-b,$ you'll get an expression that contains only even powers of $a$ and $b.$ Also, if you start with $a-b$ and multiply by $a+b,$ you'll get an expression that contains only even powers of $a$ and $b.$
One way to generalize this to a trinomial is to expand 
$$(a+b+c)(a+b-c)(a-b+c)(a-b-c)$$
By using a little strategy, this isn't as difficult as you might think:
$$[(a+b)+c][(a+b)-c]\cdot[(a-b)+c][(a-b)-c] \;\; = \;\; \left[(a+b)^2 - c^2 \right] \cdot \left[(a-b)^2 - c^2 \right]$$
$$= \;\; (a+b)^{2}(a-b)^2 - (a+b)^{2}c^2 - c^{2}(a-b)^2 + c^4$$
$$= \;\; \left[(a+b)(a-b)\right]^2 \; + \; \left[-a^2c^2 - 2abc^2 - b^2c^2 - a^2c^2 + 2abc^2 - b^2c^2 \right] + c^4$$
$$= \;\; \left(a^2 - b^2\right)^2 \; - \; 2c^2\left(a^2 + b^2\right) \; + \; c^4 $$
Notice that the result is an expression that contains only even powers of $a,$ $b,$ and $c.$ Also, notice that I didn't have to completely expand it to see that only even powers will remain.
This product of trinomials tells you that if, for example, you start with $a-b+c$ and multiply by $(a+b+c)(a+b-c)(a-b-c),$ you'll get an expression that contains only even powers of $a,$ $b,$ and $c.$ A shorthand way of saying this is that an appropriate "conjugate" of a $(-,+)$ pattern is the product of the remaining $3$ patterns. That is, an appropriate "conjugate" for a $(-,+)$ pattern is the product $(+,+)(+,-)(-,-).$
Without going into why this works (while I was writing this answer it appears that André Nicolas gave a brief, but nice, indication of how this can be done), if you start with $(a-b+c-d),$ which we might call a $(-,+,-)$ pattern, and multiply it by all of the $7$ corresponding patterns $(-,+,+),$ $(-,-,+),$ $(-,-,-),$ $(+,+,+),$ $(+,+,-),$ $(+,-,+),$ and $(+,-,-),$ you'll get an expression that contains only even powers of $a,$ $b,$ $c,$ and $d.$
The same method continues to work for additive combinations of $5$ terms, additive combinations of $6$ terms, etc. However, doing this by hand will get very tedious very soon since, for an additive combination of $5$ terms, you'll be multiplying it by $2^{4} - 1 = 15$ different expressions each consisting of $5$ terms. For an additive combination of $6$ terms, you'll be multiplying it by $2^{5} - 1 = 31$ different expressions each consisting of $6$ terms.
This method can be generalized to take care of sums in which various roots (square, cube, etc.) appear, but things quickly get even more complicated because you wind up multiplying by "$n$th roots of unity" combinations instead of sign combinations. (The signs $+$ and $-$ can be interpreted as multiplying by $1$ and by $-1,$ the two square roots of unity.) For example, radicals can be cleared from $\sqrt{a} + \sqrt[3]{b} + \sqrt{c},$ which has "sign pattern" $(+,+),$ by multiplying it by all $5$ of the "sign patterns" $(+,-),$ $(\omega,+),$ $(\omega,-),$ $({\omega}^2,+),$ and $({\omega}^2,-),$ where $\omega$ is the "first non-real cube root of $1$" as you travel counterclockwise along the unit circle in $\mathbb C$ beginning at $1$ on the positive real axis.
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