Factorise:
a3−1a3−2a+2aa^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a}a3−a31−2a+a2
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Given,
⇒a3−1a3−2a+2a⇒a3−1a3−2(a−1a)⇒(a−1a)[(a)2+1×a×1a+(1a)2]−2(a−1a)⇒(a−1a)(a2+1+1a2)−2(a−1a)⇒(a−1a)[(a2+1+1a2)−2]⇒(a−1a)(a2+1a2−1).\Rightarrow a^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a} \\[1em] \Rightarrow a^3 - \dfrac{1}{a^3} - 2\Big( a - \dfrac{1}{a}\Big) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big)\Big[(a)^2 + 1 \times a \times \dfrac{1}{a} + \Big(\dfrac{1}{a}\Big)^2\Big] - 2\Big( a - \dfrac{1}{a}\Big) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big) \Big(a^2 + 1 + \dfrac{1}{a^2}\Big)- 2\Big( a - \dfrac{1}{a}\Big) \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big) \Big[\Big(a^2 + 1 + \dfrac{1}{a^2}\Big)- 2\Big] \\[1em] \Rightarrow \Big(a - \dfrac{1}{a}\Big) \Big(a^2 + \dfrac{1}{a^2} - 1\Big).⇒a3−a31−2a+a2⇒a3−a31−2(a−a1)⇒(a−a1)[(a)2+1×a×a1+(a1)2]−2(a−a1)⇒(a−a1)(a2+1+a21)−2(a−a1)⇒(a−a1)[(a2+1+a21)−2]⇒(a−a1)(a2+a21−1).
Hence, a3−1a3−2a+2a=(a−1a)(a2+1a2−1)a^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a} = \Big(a - \dfrac{1}{a}\Big) \Big(a^2 + \dfrac{1}{a^2} - 1\Big)a3−a31−2a+a2=(a−a1)(a2+a21−1).
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