\( \left\{ \begin{array} { l } { 1.1 ) = \frac { 101 } { 100 } + \frac { 100 } { 101 } \quad b = \frac { 101 } { 100 } - \frac { 100 } { 101 } } \\ { a ^ { 2 } - b ^ { 2 } + 1 = ? } \\ { 2 x + y + z = 0 \times y z = - 18 \quad x ^ { 3 } + y ^ { 3 } + z ^ { 3 } = ? } \end{array} \right. \)
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Bonus Knowledge
To solve the first equation, we simplify \( a \) and \( b \): First, calculate \( a = \frac{101}{100} + \frac{100}{101} \). This gives us a common denominator of 10100, resulting in \( a = \frac{101^2 + 100^2}{10100} = \frac{10201 + 10000}{10100} = \frac{20201}{10100} \). Next, \( b = \frac{101}{100} - \frac{100}{101} = \frac{101^2 - 100^2}{10100} = \frac{1}{10100} \). Now, we can find \( a^2 - b^2 + 1 = (a-b)(a+b) + 1 \). For the second part, from the equations \( 2x + y + z = 0 \) and \( yz = -18 \), you can express \( z = -2x - y \). Substituting \( z \) into \( yz = -18 \) leads to a polynomial, enabling you to find \( x^3 + y^3 + z^3 \) using the identity for the sum of cubes. Keep an eye out for potential negative roots or complex numbers as they can often yield unexpected solutions!