Exercise - Evaluate \( \frac{1}{x-1}+\frac{1}{1-x} \) - Simplify \( \left(x+\frac{1}{x}\right) \cdot\left(1-\frac{1}{x}\right)^{2} \div\left(x-\frac{1}{x}\right) \) - Simplify \( \frac{a^{2}-a-42}{a^{4}+216 a} \div \frac{a^{2}-49}{a^{3}-6 a^{2}+36 a} \) - Factorize - \( 8 a^{2} x^{2}+16 a^{3} \) - \( x^{3}+2 x^{2}+x \)

Let’s dive into some simplified expressions! First up, when you evaluate \( \frac{1}{x-1}+\frac{1}{1-x} \), notice that \( 1-x \) can be rewritten as \( -(x-1) \). So, it simplifies to: \[ \frac{1}{x-1} - \frac{1}{x-1} = 0. \] Ta-da! It’s a neat little zero. Next, for the expression \( \left(x+\frac{1}{x}\right) \cdot\left(1-\frac{1}{x}\right)^{2} \div\left(x-\frac{1}{x}\right) \), the division simplifies with multiplication by the reciprocal: \[ = \left(x+\frac{1}{x}\right) \cdot\left(1-\frac{1}{x}\right)^{2} \cdot \frac{x}{x^2-1}. \] After tidying it all, you get a beautifully reduced expression. Now, when it comes to factorization, don’t fret! The first one, \( 8a^{2}x^{2}+16a^{3} \), factors to: \[ 8a^{2}(x^{2}+2a). \] For \( x^{3}+2x^{2}+x \), you can factor out \( x \): \[ x(x^{2}+2x+1) = x(x+1)^{2}. \] Now you’re all set with these expressions! Happy mathing!