Find the cube of 2a + 12a\dfrac{1}{2a}2a1, a ≠ 0;
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Solving,
(2a+12a)3=(2a)3+(12a)3+3×2a×12a×(2a+12a)⇒8a3+18a3+3(2a+12a)⇒8a3+18a3+6a+32a.\Big(2a + \dfrac{1}{2a}\Big)^3 = (2a)^3 + \Big(\dfrac{1}{2a}\Big)^3 + 3 \times 2a \times \dfrac{1}{2a} \times \Big(2a + \dfrac{1}{2a}\Big) \\[1em] \Rightarrow 8a^3 + \dfrac{1}{8a^3} + 3\Big(2a + \dfrac{1}{2a}\Big) \\[1em] \Rightarrow 8a^3 + \dfrac{1}{8a^3} + 6a + \dfrac{3}{2a}.(2a+2a1)3=(2a)3+(2a1)3+3×2a×2a1×(2a+2a1)⇒8a3+8a31+3(2a+2a1)⇒8a3+8a31+6a+2a3.
Hence, (2a+12a)3=8a3+18a3+6a+32a\Big(2a + \dfrac{1}{2a}\Big)^3 = 8a^3 + \dfrac{1}{8a^3} + 6a + \dfrac{3}{2a}(2a+2a1)3=8a3+8a31+6a+2a3.
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Find the cube of 3a−1a3a - \dfrac{1}{a}3a−a1, a ≠ 0;
If a2+1a2=47a^2 + \dfrac{1}{a^2} = 47a2+a21=47 and a ≠ 0; find :
(i) a+1aa + \dfrac{1}{a}a+a1
(ii) a3+1a3a^3 + \dfrac{1}{a^3}a3+a31
