Factorise the following:
a3−1a3−2a+2aa^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a}a3−a31−2a+a2.
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a3−1a3−2a+2a=a3−1a3−2(a−1a)a^3 - \dfrac{1}{a^3} - 2a + \dfrac{2}{a} = a^3 - \dfrac{1}{a^3} - 2\Big(a - \dfrac{1}{a}\Big)a3−a31−2a+a2=a3−a31−2(a−a1).
We know that,
a3 - b3 = (a - b)(a2 + ab + b2).
∴a3−1a3−2(a−1a)=(a−1a)(a2+a×1a+1a2)−2(a−1a)=(a−1a)(a2+1+1a2−2)=(a−1a)(a2+1a2−1).\therefore a^3 - \dfrac{1}{a^3} - 2\Big(a - \dfrac{1}{a}\Big) = \Big(a - \dfrac{1}{a}\Big)\Big(a^2 + a \times \dfrac{1}{a} + \dfrac{1}{a^2} \Big) - 2\Big(a - \dfrac{1}{a}\Big) \\[1em] = \Big(a - \dfrac{1}{a}\Big)\Big(a^2 + 1 + \dfrac{1}{a^2} - 2\Big) \\[1em] = \Big(a - \dfrac{1}{a}\Big)\Big(a^2 + \dfrac{1}{a^2} - 1\Big).∴a3−a31−2(a−a1)=(a−a1)(a2+a×a1+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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