a. p = x + ym\(^3\) ym\(^3\) = p - x m\(^3\) = \(\frac{p - x}{\text{y}}\) \(m = \sqrt[3]{\frac{p - x}{y}}\). b. x\(^2\) - 6x + 7 = 0 Using the completing the square method x\(^2\) - 6x = - 7 Add the square of half the coefficient of x to both sides x\(^2\) - 6x + (-3)\(^2\) = -7 + 9 x\(^2\) - 6x + (-3)\(^2\) = 2 (x - 3)\(^2\) = 2 x - 3 = \pm \(\sqrt{2}\) x = 3 + \(\sqrt{2}\) or 3 - \(\sqrt{2}\) x = 3 + 1.4142 or 3 - 1.4142 x = 4.1 or 1.6 to 1 decimal place. c. Let the two consecutive positive odd numbers be \(n\) and \(n + 2\). The equation for their product is: \(n(n + 2) = 195\) Expanding gives: n\(^2\) + 2n - 195 = 0 Using the quadratic formula: \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 2\), and \(c = -195\): \(b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-195) = 4 + 780 = 784\) Now substituting: \(n = \frac{-2 \pm \sqrt{784}}{2} = \frac{-2 \pm 28}{2}\) This results in: \(n = \frac{26}{2} = 13 \quad (\text{positive solution})\) Thus, the two numbers are: \(13 \quad \text{and} \quad 15\)