If a−1a=3a-\dfrac{1}{a} = 3a−a1=3; find : a2+3a+1a2−3aa^2 + 3a +\dfrac{1}{a^2} - \dfrac{3}{a}a2+3a+a21−a3 .
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Given, a−1a=3a - \dfrac{1}{a} = 3a−a1=3
Squaring both sides, we get
⇒(a−1a)2=(3)2⇒a2+1a2−2×a×1a=9⇒a2+1a2−2=9⇒a2+1a2=9+2⇒a2+1a2=11⇒ \Big(a - \dfrac{1}{a}\Big)^2 = (3)^2\\[1em] ⇒ a^2 + \dfrac{1}{a^2} - 2 \times a \times \dfrac{1}{a} = 9\\[1em] ⇒ a^2 + \dfrac{1}{a^2} - 2 = 9\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 9 + 2\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 11⇒(a−a1)2=(3)2⇒a2+a21−2×a×a1=9⇒a2+a21−2=9⇒a2+a21=9+2⇒a2+a21=11
Now,
a2+3a+1a2−3a=a2+1a2+3a−3a=a2+1a2+3(a−1a)=11+3×3=11+9=20a^2 + 3a +\dfrac{1}{a^2} - \dfrac{3}{a}\\[1em] = a^2 + \dfrac{1}{a^2} + 3a - \dfrac{3}{a}\\[1em] = a^2 + \dfrac{1}{a^2} + 3\Big(a - \dfrac{1}{a}\Big)\\[1em] = 11 + 3 \times 3\\[1em] = 11 + 9\\[1em] = 20a2+3a+a21−a3=a2+a21+3a−a3=a2+a21+3(a−a1)=11+3×3=11+9=20
Hence, a2+3a+1a2−3aa^2 + 3a +\dfrac{1}{a^2} - \dfrac{3}{a}a2+3a+a21−a3 = 20.
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