Step-up Rationalization

Using $(\sqrt{2} -1)(\sqrt{2} +1) = 1 \ and \ a^b \times\ a^c = a^{b+c} :$

$\frac{(\sqrt{2} - 1)^{1-\sqrt{3}}}{(\sqrt{2} +1)^{1+\sqrt{3}}}\times\frac{(\sqrt{2} +1)^{1-\sqrt{3}}}{(\sqrt{2} +1)^{1-\sqrt{3}}} = \frac{1}{(\sqrt{2} +1) ^2}$ $\frac{1}{(\sqrt{2} +1) ^2}\times\frac{(\sqrt{2} -1)^2}{(\sqrt{2} -1)^2} = (\sqrt{2} -1)^2 = 3 - 2\sqrt{2}$ $\huge\color{#3D99F6}{\therefore\ a+b+c = 3+2+2 = 7}$

The expression is of the following form.

$\frac{(a-b)^{c-d}}{(a+b)^{c+d}}$

We can rewrite the expression above as follows.

$\frac{(a-b)^{c}(a-b)^{-d}}{(a+b)^{c}(a+b)^{d}} = \frac{(a-b)^{c}}{(a+b)^{c}} \times \frac{1}{(a-b)^{d}(a+b)^{d}}$

=

$\frac{(a-b)^{c}}{(a+b)^{c}} \times \frac{1}{(a^2-b^2)^d} = [\frac{a-b}{a+b}]^c \times [\frac{1}{a^2 - b^2}]^d$

Now substituting $a = \sqrt{2}$ , $b = c = 1$ , and $d = \sqrt{3}$ we see that the above expression is

[\frac{\sqrt{2} - 1}{\sqrt{2} + 1}]^1 \times [\frac{1}{\sqrt{2}^2 - 1^2}]^\sqrt{3} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \times [\frac{1}{2 - 1}]^\sqrt{3}

=

\frac{\sqrt{2} - 1}{\sqrt{2} + 1} \times 1^\sqrt{3} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1}

Rationalizing the denominator gives us that the above expression is equal to

$(\sqrt{2} - 1)^2 = 3 - 2\sqrt{2}$

Finally we see that 3 + 2 + 2 = 7!

Curtis's solution is much more slick. lol.

This an easy problem. My solution wasn't as simple as Curtis Clement's, but it gets the job done. :)

The top part, (sqrt2-1)^(1-sqrt3) can easily be reincorporated into the denominator. Since 1-sqrt3 is a negative in the numerator, we can apply it in to the denominator with the exponent being sqrt3-1.

We get this: 1 -OVER- (sqrt2 - 1)^(sqrt3 - 1) * (sqrt2+1)^(1+sqrt3)

We try to rationalize the exponent in the (sqrt2 - 1)^(sqrt3 - 1) part by raising this whole fraction to the power of sqrt3 + 1 (or 1+sqrt3, hang on, this is going to be important).

We get this: 1 -OVER- (sqrt2-1)^(2) * (sqrt2+1)^(4+2sqrt3)

(The exponent two came from difference of squares [1-sqrt3] * [1+sqrt3] and the 4+2sqrt3 exponent came from (a+b)(a+b) = a^2 +2ab+ b^2)

We draw out from the (sqrt2+1)^(4+2sqrt3) part and separate into

(sqrt2+1)^(2) * (sqrt2+1)^(2+2sqrt3) (exponential rules)

Again, from here, we can use difference of squares since

(sqrt2-1)^(2) and (sqrt2+1)^(2) can be reorganized as [(sqrt2-1)(sqrt2+1)]^2

AKA 1.

We get this: 1 -OVER- 1 * (sqrt2+1)^(2+2sqrt3)

Now from here, we can again separate (sqrt2+1)^(2+2sqrt3) into

[(sqrt2+1)^(2)] ^ (1+sqrt3)

Since we have originally raised the original fraction to (1+sqrt3)th power, it makes sense to take the (1+sqrt3)th root to get the original fraction back.

1 to any root or power is 1, and taking the (1 + sqrt3)th root of the denominator results in just (sqrt2+1)^2

Rationalizing the root from there, we get (sqrt2-1)^2

Which is equal to 3-2sqrt2

In the form of a-bsqrtc, a+b+c = 3+2+2 = 7.

And we are done.