<p>If <em>ax</em><sup>2</sup> + <em>bx</em> + <em>c</em> = 0 and <em>bx</em><sup>2</sup> + <em>cx</em> + <em>a</em> = 0 have a common root and <em>a</em>, <em>b</em>, and <em>c</em> are nonzero real numbers, then find the value of \(\dfrac{a^3 + b^3 + c^3}{abc}\).</p>
Step-by-Step Solution
Key Concept: If two quadratic equations share a common root, we can use the relationship between coefficients by eliminating that root. Dividing the first equation by x² and comparing with the second equation reveals that a³ + b³ + c³ = 3abc.
<p><strong>Step 1:</strong> Let r be the common root. Then:</p><p>ar² + br + c = 0 ... (1)</p><p>br² + cr + a = 0 ... (2)</p><p><strong>Step 2:</strong> Divide equation (1) by r² (since r ≠ 0, as a, b, c are nonzero):</p><p>a + b/r + c/r² = 0</p><p>Multiply by r²: ar² + br + c = 0 ✓</p><p><strong>Step 3:</strong> Now, rearrange as: c/r² + b/r + a = 0</p><p>This means if r is a root of ax² + bx + c = 0, then 1/r is a root of cx² + bx + a = 0.</p><p><strong>Step 4:</strong> Since r satisfies both ax² + bx + c = 0 and bx² + cx + a = 0, and 1/r satisfies cx² + bx + a = 0, we must have r = 1/r, giving r² = 1, so r = ±1.</p><p><strong>Step 5:</strong> Test r = 1 in both equations:</p><p>From (1): a + b + c = 0</p><p>From (2): b + c + a = 0 ✓</p><p>This is consistent!</p><p><strong>Step 6:</strong> If a + b + c = 0, then c = -(a + b).</p><p>Using the identity: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)</p><p>Since a + b + c = 0:</p><p>a³ + b³ + c³ - 3abc = 0</p><p>∴ a³ + b³ + c³ = 3abc</p><p><strong>∴ Answer:</strong> (a³ + b³ + c³)/abc = 3</p>
Correct Answer: 3