Simplify the following:
(a+1a)2−(a−1a)2\Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2(a+a1)2−(a−a1)2
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(a+1a)2−(a−1a)2=[(a)2+2(a)(1a)+(1a)2]−[(a)2−2(a)(1a)+(1a)2]=[a2+2+1a2]−[a2−2+1a2]=a2+2+1a2−a2+2−1a2=4\Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = \Big[\Big(a\Big)^2 + 2\Big(a\Big)\Big(\dfrac{1}{a}\Big) + \Big(\dfrac{1}{a}\Big)^2 \Big] - \Big[\Big(a\Big)^2 - 2\Big(a\Big)\Big(\dfrac{1}{a}\Big) + \Big(\dfrac{1}{a}\Big)^2\Big] \\[1em] = \Big[a^2 + 2 + \dfrac{1}{a^2} \Big] - \Big[a^2 - 2 + \dfrac{1}{a^2} \Big] \\[1em] = a^2 + 2 + \dfrac{1}{a^2} - a^2 + 2 - \dfrac{1}{a^2} = 4(a+a1)2−(a−a1)2=[(a)2+2(a)(a1)+(a1)2]−[(a)2−2(a)(a1)+(a1)2]=[a2+2+a21]−[a2−2+a21]=a2+2+a21−a2+2−a21=4
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By using standard formulae, expand the following:
(2x−13y)3\Big(2x - \dfrac{1}{3y}\Big)^3(2x−3y1)3
(a+1a)2+(a−1a)2\Big(a + \dfrac{1}{a}\Big)^2 + \Big(a - \dfrac{1}{a}\Big)^2(a+a1)2+(a−a1)2
(3x-1)2 - (3x-2)(3x+1)
(4x+3y)2 - (4x-3y)2 - 48xy
