Find : a3−1a3a^3 -\dfrac{1}{a^3}a3−a31, if a−1a=4a -\dfrac{1}{a}= 4a−a1=4.
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Using the formula,
[∵ (x - y)3 = x3 - 3xy(x - y) - y3]
So,
(a−1a)3=a3−3×a×1a(a−1a)−(1a)3⇒(a−1a)3=a3−3(a−1a)−1a3\Big(a - \dfrac{1}{a}\Big)^3 = a^3 - 3 \times a \times \dfrac{1}{a}\Big(a - \dfrac{1}{a}\Big) - \Big(\dfrac{1}{a}\Big)^3\\[1em] ⇒ \Big(a - \dfrac{1}{a}\Big)^3 = a^3 - 3\Big(a - \dfrac{1}{a}\Big) - \dfrac{1}{a^3}(a−a1)3=a3−3×a×a1(a−a1)−(a1)3⇒(a−a1)3=a3−3(a−a1)−a31
Putting a−1a=4a - \dfrac{1}{a} = 4a−a1=4
43=a3−3×4−1a3⇒64=a3−12−1a3⇒a3−1a3=64+12⇒a3−1a3=764^3 = a^3 - 3 \times 4 - \dfrac{1}{a^3}\\[1em] ⇒ 64 = a^3 - 12 - \dfrac{1}{a^3}\\[1em] ⇒ a^3 - \dfrac{1}{a^3} = 64 + 12\\[1em] ⇒ a^3 - \dfrac{1}{a^3} = 7643=a3−3×4−a31⇒64=a3−12−a31⇒a3−a31=64+12⇒a3−a31=76
Hence, the value of a3−1a3a^3 - \dfrac{1}{a^3}a3−a31 is 76.
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