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(a+1a)3\Big(a +\dfrac{1}{a}\Big)^3(a+a1)3
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Using the formula,
(x + y)3 = x3 + y3 + 3x2y + 3xy2
=(a)3+(1a)3+3×(a)2×(1a)+3×(a)×(1a)2=a3+(1a3)+(3a2a)+(3aa2)=a3+(1a3)+3a+(3a)= (a)^3 + \Big(\dfrac{1}{a}\Big)^3 + 3 \times (a)^2 \times \Big(\dfrac{1}{a}\Big) + 3 \times (a) \times \Big(\dfrac{1}{a}\Big)^2\\[1em] = a^3 + \Big(\dfrac{1}{a^3}\Big) + \Big(\dfrac{3a^2}{a}\Big) + \Big(\dfrac{3a}{a^2}\Big)\\[1em] = a^3 + \Big(\dfrac{1}{a^3}\Big) + 3a + \Big(\dfrac{3}{a}\Big)\\[1em]=(a)3+(a1)3+3×(a)2×(a1)+3×(a)×(a1)2=a3+(a31)+(a3a2)+(a23a)=a3+(a31)+3a+(a3)
Hence, (a+1a)3=a3+(1a3)+3a+(3a)\Big(a +\dfrac{1}{a}\Big)^3 = a^3 + \Big(\dfrac{1}{a^3}\Big) + 3a + \Big(\dfrac{3}{a}\Big)(a+a1)3=a3+(a31)+3a+(a3)
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