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(2a−12a)3\Big(2a -\dfrac{1}{2a}\Big)^3(2a−2a1)3
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Using the formula,
(x - y)3 = x3 - y3 - 3x2y + 3xy2
=(2a)3−(12a)3−3×(2a)2×(12a)+3×(2a)×(12a)2=8a3−(18a3)−(12a22a)+(6a4a2)=8a3−(18a3)−6a+(32a)= (2a)^3 - \Big(\dfrac{1}{2a}\Big)^3 - 3 \times (2a)^2 \times \Big(\dfrac{1}{2a}\Big) + 3 \times (2a) \times \Big(\dfrac{1}{2a}\Big)^2\\[1em] = 8a^3 - \Big(\dfrac{1}{8a^3}\Big) - \Big(\dfrac{12a^2}{2a}\Big) + \Big(\dfrac{6a}{4a^2}\Big)\\[1em] = 8a^3 - \Big(\dfrac{1}{8a^3}\Big) - 6a + \Big(\dfrac{3}{2a}\Big)\\[1em]=(2a)3−(2a1)3−3×(2a)2×(2a1)+3×(2a)×(2a1)2=8a3−(8a31)−(2a12a2)+(4a26a)=8a3−(8a31)−6a+(2a3)
Hence, (2a−12a)3=8a3−(18a3)−6a+(32a)\Big(2a - \dfrac{1}{2a}\Big)^3 = 8a^3 - \Big(\dfrac{1}{8a^3}\Big) - 6a + \Big(\dfrac{3}{2a}\Big)(2a−2a1)3=8a3−(8a31)−6a+(2a3)
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(x + 5y)3
(a+1a)3\Big(a +\dfrac{1}{a}\Big)^3(a+a1)3
Find the cube of:
a + 2
2a - 1
