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]=2a2+2a2=2(a2+1a2)\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] = 2a^2 + \dfrac{2}{a^2} \\[1em] = 2\Big(a^2 + \dfrac{1}{a^2}\Big)(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]=2a2+a22=2(a2+a21)
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By using standard formulae, expand the following:
(5x-3y)3
(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)
