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If a+1a=2a + \dfrac{1}{a} = 2; find :

(i) a4+1a2\dfrac{a^4 + 1}{a^2}

(ii) a8+1a4\dfrac{a^8 + 1}{a^4}

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Answer

(i) Given, a+1a=2a + \dfrac{1}{a} = 2

Squaring both sides, we get

⇒(a+1a)2=(2)2⇒a2+1a2+2×a×1a=4⇒a2+1a2+2=4⇒a2+1a2=4−2⇒a2+1a2=2⇒a4+1a2=2⇒ \Big(a + \dfrac{1}{a}\Big)^2 = (2)^2\\[1em] ⇒ a^2 + \dfrac{1}{a^2} + 2 \times a \times \dfrac{1}{a} = 4\\[1em] ⇒ a^2 + \dfrac{1}{a^2} + 2 = 4\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 4 - 2\\[1em] ⇒ a^2 + \dfrac{1}{a^2} = 2\\[1em] ⇒ \dfrac{a^4 + 1}{a^2} = 2\\[1em]

Hence, a4+1a2\dfrac{a^4 + 1}{a^2} = 2.

(ii) a2+1a2=2a^2 + \dfrac{1}{a^2} = 2

Squaring both sides, we get

⇒(a2+1a2)2=(2)2⇒a4+1a4+2×a2×1a2=4⇒a4+1a4+2=4⇒a4+1a4=4−2⇒a4+1a4=2⇒a8+1a4=2⇒ \Big(a^2 + \dfrac{1}{a^2}\Big)^2 = (2)^2\\[1em] ⇒ a^4 + \dfrac{1}{a^4} + 2 \times a^2 \times \dfrac{1}{a^2} = 4\\[1em] ⇒ a^4 + \dfrac{1}{a^4} + 2 = 4\\[1em] ⇒ a^4 + \dfrac{1}{a^4} = 4 - 2\\[1em] ⇒ a^4 + \dfrac{1}{a^4} = 2\\[1em] ⇒ \dfrac{a^8 + 1}{a^4} = 2\\[1em]

Hence, a8+1a4\dfrac{a^8 + 1}{a^4} = 2.

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