By using standard formulae, expand the following:
(2x+3x−1)2\Big(2x + \dfrac{3}{x} - 1\Big)^2(2x+x3−1)2
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(2x+3x−1)2=[(2x)+(3x)+(−1)]2=(2x)2+(3x)2+(−1)2+2[(2x)(3x)+(3x)(−1)+(−1)(2x)]=4x2+9x2+1+2[6−3x−2x]=4x2+9x2+1+12−6x−4x=4x2+9x2+13−6x−4x\Big(2x + \dfrac{3}{x} - 1\Big)^2 = \Big[\Big(2x\Big) + \Big(\dfrac{3}{x}\Big) + \Big(-1\Big)\Big]^2 \\[1em] = (2x)^2 + \Big(\dfrac{3}{x}\Big)^2 + (-1)^2 + 2 \Big[\Big(2x\Big)\Big(\dfrac{3}{x}\Big) + \Big(\dfrac{3}{x}\Big)\Big(-1\Big)+\Big(-1\Big)\Big(2x\Big) \Big] \\[1em] = 4x^2 + \dfrac{9}{x^2} + 1 + 2\Big[6 - \dfrac{3}{x} - 2x \Big] \\[1em] = 4x^2 + \dfrac{9}{x^2} + 1 + 12 - \dfrac{6}{x} - 4x \\[1em] = 4x^2 + \dfrac{9}{x^2} + 13 - \dfrac{6}{x} - 4x \\[1em](2x+x3−1)2=[(2x)+(x3)+(−1)]2=(2x)2+(x3)2+(−1)2+2[(2x)(x3)+(x3)(−1)+(−1)(2x)]=4x2+x29+1+2[6−x3−2x]=4x2+x29+1+12−x6−4x=4x2+x29+13−x6−4x
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(x - 2y - z)2
(2x - 3y + 4z)2
(23x−32x−1)2\Big(\dfrac{2}{3}x - \dfrac{3}{2x} - 1\Big)^2(32x−2x3−1)2
(x+2)3
