Quadratic Equations
Algebraic identities
Grade 11
Question:
<p>If \( a + b + c = 0 \), then the value of \((x-1)^3 + (2x-1)^3 + (2-3x)^3\) equals:</p>
<p>A) \( 3(x-1)(2x-1)(2-3x) \)</p>
<p>B) \( -3(x-1)(2x-1)(3x-2) \)</p>
<p>C) \( 3(x-1)(2x-1)(3x-2) \)</p>
<p>D) \( -(x-1)(2x-1)(2-3x) \)</p>
Step-by-Step Solution
Key Concept: When a + b + c = 0, the identity a³ + b³ + c³ = 3abc applies directly. Recognize that (x-1) + (2x-1) + (2-3x) = 0, making this the perfect setup for the identity.
<p><strong>Step 1:</strong> Verify the sum condition. Let a = (x-1), b = (2x-1), c = (2-3x).</p><p>Calculate: a + b + c = (x-1) + (2x-1) + (2-3x) = x + 2x - 3x - 1 - 1 + 2 = 0 ✓</p><p><strong>Step 2:</strong> Apply the identity a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).</p><p>Since a + b + c = 0, we have: a³ + b³ + c³ = 3abc</p><p><strong>Step 3:</strong> Calculate abc = (x-1)(2x-1)(2-3x).</p><p>First: (x-1)(2x-1) = 2x² - 3x + 1</p><p>Then: (2x² - 3x + 1)(2-3x) = 4x² - 6x³ - 6x + 9x² + 2 - 3x = -6x³ + 13x² - 9x + 2</p><p><strong>Step 4:</strong> Therefore: (x-1)³ + (2x-1)³ + (2-3x)³ = 3(-6x³ + 13x² - 9x + 2) = <strong>-18x³ + 39x² - 27x + 6</strong></p><p>Or factored: <strong>3(2-3x)(2x² - 3x + 1)</strong> or simplified to <strong>3(x-1)(2x-1)(2-3x)</strong> × 3</p><p>∴ Answer: A</p>
Correct Answer: A