<p>Let \(r\), \(s\), and \(t\) be the roots of equation \(8x^3 + 1001x + 2008 = 0\). Then find the value of \((r+s)^3 + (s+t)^3 + (t+r)^3\).</p>
Step-by-Step Solution
Key Concept: Since r, s, t are roots of the cubic, use Vieta's formulas to find r+s = -t, s+t = -r, and t+r = -s. Then the sum of cubes becomes (-t)³ + (-r)³ + (-s)³ = -(r³+s³+t³), which can be evaluated using the identity r³+s³+t³-3rst = (r+s+t)(r²+s²+t²-rs-st-tr).
<p><strong>Step 1: Apply Vieta's Formulas</strong></p><p>For 8x³ + 1001x + 2008 = 0, or x³ + (1001/8)x + 251 = 0:</p><p>• r + s + t = 0 (coefficient of x² is 0)</p><p>• rs + st + tr = 1001/8</p><p>• rst = -251</p><p><strong>Step 2: Use the constraint r + s + t = 0</strong></p><p>Since r + s + t = 0:</p><p>• r + s = -t</p><p>• s + t = -r</p><p>• t + r = -s</p><p><strong>Step 3: Substitute into the expression</strong></p><p>(r+s)³ + (s+t)³ + (t+r)³ = (-t)³ + (-r)³ + (-s)³</p><p>= -t³ - r³ - s³ = -(r³ + s³ + t³)</p><p><strong>Step 4: Apply the cubic identity</strong></p><p>r³ + s³ + t³ - 3rst = (r+s+t)(r² + s² + t² - rs - st - tr)</p><p>Since r + s + t = 0:</p><p>r³ + s³ + t³ - 3rst = 0</p><p>∴ r³ + s³ + t³ = 3rst = 3(-251) = -753</p><p><strong>Step 5: Final calculation</strong></p><p>(r+s)³ + (s+t)³ + (t+r)³ = -(r³ + s³ + t³) = -(-753) = 753</p><p>∴ Answer: <strong>753</strong></p>
Correct Answer: 753