<p>The term independent of \(x\) in expansion of \(\left(\dfrac{x+1}{x^{2/3} - x^{1/3} + 1} - \dfrac{x-1}{x - x^{1/2}}\right)^{10}\) is</p>
Step-by-Step Solution
Key Concept: Simplify each fraction by recognizing algebraic identities: the first uses a³+b³ factorization and the second uses difference of squares, then combine to find the base expression before applying binomial expansion.
<p><strong>Step 1: Simplify the first fraction</strong></p><p>Note that x²/³ - x^(1/3) + 1 is of the form a² - ab + b² where a = x^(1/3) and b = 1.</p><p>Since (a+b)(a²-ab+b²) = a³+b³, we have:</p><p>(x^(1/3)+1)(x^(2/3)-x^(1/3)+1) = x + 1</p><p>Therefore: (x+1)/(x^(2/3)-x^(1/3)+1) = x^(1/3)+1</p><p><strong>Step 2: Simplify the second fraction</strong></p><p>For the second fraction: (x-1)/(x-x^(1/2)) = (x-1)/(x^(1/2)(x^(1/2)-1))</p><p>= (x^(1/2)-1)(x^(1/2)+1)/(x^(1/2)(x^(1/2)-1)) = (x^(1/2)+1)/x^(1/2) = 1 + x^(-1/2)</p><p><strong>Step 3: Combine the expressions</strong></p><p>The expression becomes: [(x^(1/3)+1) - (1+x^(-1/2))]^(10) = [x^(1/3) - x^(-1/2)]^(10)</p><p><strong>Step 4: Find the independent term using binomial theorem</strong></p><p>General term: T_(r+1) = C(10,r)(x^(1/3))^(10-r)(-x^(-1/2))^r = C(10,r)(-1)^r · x^((10-r)/3 - r/2)</p><p>For independence: (10-r)/3 - r/2 = 0</p><p>⟹ 2(10-r) - 3r = 0 ⟹ 20 - 5r = 0 ⟹ r = 4</p><p><strong>Step 5: Calculate the coefficient</strong></p><p>T₅ = C(10,4)(-1)⁴ = 210 · 1 = 210</p><p>∴ Answer: C</p>
Correct Answer: C