Simplify (b − 1)/(b + √b − 1) times the reciprocal of (√b + 1)/(b√b − 1), then add 2√b.

Step-by-step solution

1. 1

   Clear the negative exponent by flipping the fraction. For any nonzero fraction, $\left(\tfrac{u}{v}\right)^{-1}=\tfrac{v}{u}$, so

   $\left(\frac{\sqrt{b}+1}{b\sqrt{b}-1}\right)^{-1}=\frac{b\sqrt{b}-1}{\sqrt{b}+1}$

   and the expression becomes a plain product plus a term:

   $\frac{b-1}{b+\sqrt{b}-1}\cdot\frac{b\sqrt{b}-1}{\sqrt{b}+1}+2\sqrt{b}$

2. 2

   Substitute $x=\sqrt{b}$ so that everything is a polynomial. This is the key move: radicals block the usual factoring identities, but with $x=\sqrt{b}\ (x>0)$ we get $b=x^2$ and $b\sqrt{b}=x^3$, turning the expression into

   $\frac{x^2-1}{x^2+x-1}\cdot\frac{x^3-1}{x+1}+2x$

3. 3

   Factor and cancel. Now the standard identities apply:

   $x^2-1=(x-1)(x+1),\qquad x^3-1=(x-1)(x^2+x+1)$

   so the product is

   $\frac{(x-1)(x+1)}{x^2+x-1}\cdot\frac{(x-1)(x^2+x+1)}{x+1}=\frac{(x-1)^2(x^2+x+1)}{x^2+x-1}$

   The $x+1$ cancels; note $x^2+x-1$ (the denominator) is not the same as $x^2+x+1$, so nothing else cancels.

4. 4

   Add $2x$ over the common denominator. Write $2x=\dfrac{2x(x^2+x-1)}{x^2+x-1}$ and combine:

   $\frac{(x-1)^2(x^2+x+1)+2x(x^2+x-1)}{x^2+x-1}$

   Expand the numerator:

   $(x^2-2x+1)(x^2+x+1)=x^4-x^3-x+1$

   $2x(x^2+x-1)=2x^3+2x^2-2x$

   $\text{sum}=x^4+x^3+2x^2-3x+1$

5. 5

   Check that the fraction really is in lowest terms. Dividing $x^4+x^3+2x^2-3x+1$ by $x^2+x-1$ leaves quotient $x^2+3$ and remainder $-6x+4\neq 0$, so the division does not come out even and no further cancellation is possible.

6. 6

   Substitute back and state the domain. With $x^2=b$, $x^3=b\sqrt{b}$, $x^4=b^2$:

   $\frac{b^2+b\sqrt{b}+2b-3\sqrt{b}+1}{b+\sqrt{b}-1}$

   This is valid for $b>0$ with $b\neq 1$ (so $b\sqrt{b}-1\neq 0$) and $b\neq\frac{3-\sqrt{5}}{2}\approx 0.382$ (so $b+\sqrt{b}-1\neq 0$). A numeric check at $b=2$ gives $3.142136$ from both the original and the simplified form.