Find the cube of 3a−1a3a - \dfrac{1}{a}3a−a1, a ≠ 0;
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Solving,
⇒(3a−1a)3=(3a)3−(1a)3−3×3a×1a×(3a−1a)=27a3−1a3−9(3a−1a)=27a3−1a3−27a+9a.\Rightarrow \Big(3a - \dfrac{1}{a}\Big)^3 = (3a)^3 - \Big(\dfrac{1}{a}\Big)^3 - 3 \times 3a \times \dfrac{1}{a} \times \Big(3a - \dfrac{1}{a}\Big) \\[1em] = 27a^3 - \dfrac{1}{a^3} - 9\Big(3a - \dfrac{1}{a}\Big) \\[1em] = 27a^3 - \dfrac{1}{a^3} - 27a + \dfrac{9}{a}.⇒(3a−a1)3=(3a)3−(a1)3−3×3a×a1×(3a−a1)=27a3−a31−9(3a−a1)=27a3−a31−27a+a9.
Hence, (3a−1a)3=27a3−1a3−27a+9a.\Big(3a - \dfrac{1}{a}\Big)^3 = 27a^3 - \dfrac{1}{a^3} - 27a + \dfrac{9}{a}.(3a−a1)3=27a3−a31−27a+a9.
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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
If a2+1a2=18a^2 + \dfrac{1}{a^2} = 18a2+a21=18 and a ≠ 0; find :
(i) a−1aa - \dfrac{1}{a}a−a1
(ii) a3−1a3a^3 - \dfrac{1}{a^3}a3−a31
