By using standard formulae, expand the following:
(3x+1x)3\Big(3x + \dfrac{1}{x}\Big)^3(3x+x1)3
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(3x+1x)3=(3x)3+(1x)3+3(3x)(1x)(3x+1x)=27x3+1x3+27x+9x\Big(3x + \dfrac{1}{x}\Big)^3 = (3x)^3 + \Big(\dfrac{1}{x}\Big)^3 + 3(3x)\Big(\dfrac{1}{x}\Big)\Big(3x + \dfrac{1}{x}\Big) \\[1em] = 27x^3 + \dfrac{1}{x^3} + 27x + \dfrac{9}{x}(3x+x1)3=(3x)3+(x1)3+3(3x)(x1)(3x+x1)=27x3+x31+27x+x9
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(x+2)3
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