<p>Value of \(\dfrac{(x-1)^3 + (2x-1)^3 - (3x-2)^3}{(x-1)(2x-1)(3x-2)}\) is equal to:</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator has the form a³ + b³ + c³ where a + b + c = 0, which triggers the identity a³ + b³ + c³ = 3abc when the sum equals zero.
<p><strong>Step 1:</strong> Check if the sum of terms in numerator equals zero.</p><p>Let a = (x-1), b = (2x-1), c = -(3x-2)</p><p>Then: (x-1) + (2x-1) + (-(3x-2)) = x - 1 + 2x - 1 - 3x + 2 = 0 ✓</p><p><strong>Step 2:</strong> Apply the identity: When a + b + c = 0, then a³ + b³ + c³ = 3abc</p><p>Therefore: (x-1)³ + (2x-1)³ + (-(3x-2))³ = 3(x-1)(2x-1)(-(3x-2))</p><p><strong>Step 3:</strong> Rewrite the original expression:</p><p>(x-1)³ + (2x-1)³ - (3x-2)³ = 3(x-1)(2x-1)(-(3x-2)) = -3(x-1)(2x-1)(3x-2)</p><p><strong>Step 4:</strong> Divide by denominator:</p><p>∴ Answer: -3(x-1)(2x-1)(3x-2) / [(x-1)(2x-1)(3x-2)] = <strong>-3</strong></p>
Correct Answer: A