Question

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$(2x-\frac{1}{x}{)}^6$

Answer 1

$(2x-\frac{1}{x}{)}^6$

${\Rightarrow }^6{C}_0\left(2x{)}^6{+}^6{C}_1\right(2x{)}^5\left(\frac{-1}{x}\right){+}^6{C}_2\left(2x{)}^x\right(-\frac{1}{x}{)}^x{+}^6{C}_3\left(2x{)}^3\right(\frac{-1}{x}{)}^3$

${+}^6{C}_4\left(2x{)}^1\right(\frac{1}{x}{)}^4{+}^6{C}_5\left(2x\right)\left(\frac{-1}{x}{)}^5{+}^6{C}_6\right(\frac{-1}{x}{)}^6$

$\Rightarrow 1(2x{)}^6+6(2x{)}^5\left(\frac{-1}{x}\right)+\frac{6\times 5\times (22{)}^4}{2}(\frac{-1}{x}{)}^2+$

${}^6{C}_3\frac{6\times 5\times 4}{3\times 2}\left(2x{)}^3\right(\frac{-1}{x}{)}^3+\frac{6\times 5}{2}\times \left(2x{)}^3\right(\frac{-1}{x}{)}^4+6\left(2x\right)(\frac{-1}{x}{)}^5+1(\frac{-1}{x}{)}^6$

$\Rightarrow 64x^6+{62}^5{x}^5\left(\frac{-1}{x}\right)+15.16\frac{x^4}{x^2}+20\left(8\right)\frac{-x^3}{x^3}+{5.2}^2\frac{x^2}{x^4}+12\left(\frac{-x}{x^5}\right)+\frac{1}{x^6}$

$\Rightarrow 64x^6-{62}^5{x}^4+240x^2-160+\frac{20}{x^2}-\frac{12}{x^4}+\frac{1}{x_6}$

$64x^6-192x^4+240x^2-160+\frac{20}{x^2}-\frac{12}{x^4}+\frac{1}{x^6}$