By using standard formulae, expand the following:
(23x−32x−1)2\Big(\dfrac{2}{3}x - \dfrac{3}{2x} - 1\Big)^2(32x−2x3−1)2
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(23x−32x−1)2=[23x+(−32x)+(−1)]2=(23x)2+(−32x)2+(−1)2+2[(23x)(−32x)+(−32x)(−1)+(−1)(23x)]=49x2+94x2+1+2[−1+32x−23x]=49x2+94x2+1−2+3x−43x=49x2+94x2−1+3x−43x\Big(\dfrac{2}{3}x - \dfrac{3}{2x} - 1\Big)^2 = \Big[\dfrac{2}{3}x + \Big(-\dfrac{3}{2x}\Big) + \Big(-1\Big)\Big]^2 \\[1em] = \Big(\dfrac{2}{3}x\Big)^2 + \Big(-\dfrac{3}{2x}\Big)^2 + (-1)^2 + 2 \Big[\Big(\dfrac{2}{3}x\Big)\Big(-\dfrac{3}{2x}\Big) + \Big(-\dfrac{3}{2x}\Big)\Big(-1\Big)+\Big(-1\Big)\Big(\dfrac{2}{3}x\Big) \Big] \\[1em] = \dfrac{4}{9}x^2 + \dfrac{9}{4x^2} + 1 + 2\Big[-1 + \dfrac{3}{2x} - \dfrac{2}{3}x \Big] \\[1em] = \dfrac{4}{9}x^2 + \dfrac{9}{4x^2} + 1 - 2 + \dfrac{3}{x} - \dfrac{4}{3}x \\[1em] = \dfrac{4}{9}x^2 + \dfrac{9}{4x^2} -1 + \dfrac{3}{x} - \dfrac{4}{3}x \\[1em](32x−2x3−1)2=[32x+(−2x3)+(−1)]2=(32x)2+(−2x3)2+(−1)2+2[(32x)(−2x3)+(−2x3)(−1)+(−1)(32x)]=94x2+4x29+1+2[−1+2x3−32x]=94x2+4x29+1−2+x3−34x=94x2+4x29−1+x3−34x
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(2x - 3y + 4z)2
(2x+3x−1)2\Big(2x + \dfrac{3}{x} - 1\Big)^2(2x+x3−1)2
(x+2)3
(2a+b)3
