5. Complete the following. a) Simplify: i) \( \left(m+\frac{1}{m}\right)\left(m^{2}-1+\frac{1}{m^{2}}\right) \) ii) \( \left(m+\frac{1}{m}\right)^{2} \). b) If \( m+\frac{1}{m}=2 \), calculate the value of: i) \( m^{2}+\frac{1}{m^{2}} \) ii) \( m^{3}+\frac{1}{m^{3}} \)

Certainly! Let's tackle each part of the problem step-by-step. --- ## **Problem 5** ### **a) Simplify:** #### **i) Simplify \( \left(m + \frac{1}{m}\right) \left(m^{2} - 1 + \frac{1}{m^{2}}\right) \)** **Solution:** To simplify the expression, we'll multiply the two binomials term by term. \[ \left(m + \frac{1}{m}\right) \left(m^{2} - 1 + \frac{1}{m^{2}}\right) \] **Step 1:** Expand the product using the distributive property (FOIL method): \[ = m \cdot m^{2} + m \cdot (-1) + m \cdot \frac{1}{m^{2}} + \frac{1}{m} \cdot m^{2} + \frac{1}{m} \cdot (-1) + \frac{1}{m} \cdot \frac{1}{m^{2}} \] **Step 2:** Multiply each term: \[ = m^3 - m + \frac{m}{m^{2}} + \frac{m^{2}}{m} - \frac{1}{m} + \frac{1}{m^{3}} \] Simplify the fractions: \[ = m^3 - m + \frac{1}{m} + m - \frac{1}{m} + \frac{1}{m^{3}} \] **Step 3:** Combine like terms: \[ - m + m = 0 \quad \text{and} \quad \frac{1}{m} - \frac{1}{m} = 0 \] So, the expression simplifies to: \[ m^3 + \frac{1}{m^{3}} \] **Final Answer:** \[ \boxed{m^{3} + \dfrac{1}{m^{3}}} \] --- #### **ii) Simplify \( \left(m + \frac{1}{m}\right)^{2} \)** **Solution:** We'll use the formula for squaring a binomial: \[ \left(m + \frac{1}{m}\right)^{2} = m^{2} + 2 \cdot m \cdot \frac{1}{m} + \left(\frac{1}{m}\right)^{2} \] **Step 1:** Simplify each term: \[ = m^{2} + 2 \cdot 1 + \frac{1}{m^{2}} \] \[ = m^{2} + 2 + \frac{1}{m^{2}} \] **Final Answer:** \[ \boxed{m^{2} + 2 + \dfrac{1}{m^{2}}} \] --- ### **b) If \( m + \frac{1}{m} = 2 \), calculate the value of:** #### **i) \( m^{2} + \frac{1}{m^{2}} \)** **Solution:** Given: \[ m + \frac{1}{m} = 2 \] We'll square both sides to find \( m^{2} + \frac{1}{m^{2}} \). **Step 1:** Square both sides: \[ \left(m + \frac{1}{m}\right)^{2} = 2^{2} \] \[ m^{2} + 2 \cdot m \cdot \frac{1}{m} + \frac{1}{m^{2}} = 4 \] Simplify: \[ m^{2} + 2 + \frac{1}{m^{2}} = 4 \] **Step 2:** Subtract 2 from both sides: \[ m^{2} + \frac{1}{m^{2}} = 4 - 2 = 2 \] **Final Answer:** \[ \boxed{2} \] --- #### **ii) \( m^{3} + \frac{1}{m^{3}} \)** **Solution:** We can use the identity for the cube of a binomial: \[ \left(m + \frac{1}{m}\right)^{3} = m^{3} + 3m + 3\left(\frac{1}{m}\right) + \frac{1}{m^{3}} \] Which simplifies to: \[ m^{3} + \frac{1}{m^{3}} + 3\left(m + \frac{1}{m}\right) \] Given \( m + \frac{1}{m} = 2 \), we can rearrange the formula to solve for \( m^{3} + \frac{1}{m^{3}} \): \[ m^{3} + \frac{1}{m^{3}} = \left(m + \frac{1}{m}\right)^{3} - 3\left(m + \frac{1}{m}\right) \] **Step 1:** Compute \( \left(m + \frac{1}{m}\right)^{3} \): \[ 2^{3} = 8 \] **Step 2:** Multiply 3 by \( m + \frac{1}{m} \): \[ 3 \times 2 = 6 \] **Step 3:** Subtract: \[ m^{3} + \frac{1}{m^{3}} = 8 - 6 = 2 \] **Final Answer:** \[ \boxed{2} \] ---