By using standard formulae, expand the following:
(2x−13y)3\Big(2x - \dfrac{1}{3y}\Big)^3(2x−3y1)3
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(2x−13y)3=(2x)3−(13y)3−3(2x)(13y)(2x−13y)=8x3−127y3−4x2y+2x3y2\Big(2x - \dfrac{1}{3y}\Big)^3 = (2x)^3 - \Big(\dfrac{1}{3y}\Big)^3 - 3(2x)\Big(\dfrac{1}{3y}\Big)\Big(2x - \dfrac{1}{3y}\Big) \\[1em] = 8x^3 - \dfrac{1}{27y^3} - \dfrac{4x^2}{y} + \dfrac{2x}{3y^2}(2x−3y1)3=(2x)3−(3y1)3−3(2x)(3y1)(2x−3y1)=8x3−27y31−y4x2+3y22x
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(2x-1)3
(5x-3y)3
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
(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
