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Mathematics

If a−1a=5a - \dfrac{1}{a} = \sqrt{5}, find the values of :

(i) (a+1a)\Big(a + \dfrac{1}{a}\Big)

(ii) (a3+1a3)\Big(a^3 + \dfrac{1}{a^3}\Big)

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Answer

(i) Given,

a−1a=5a - \dfrac{1}{a} = \sqrt{5}

Using identity,

⇒(a+1a)2−(a−1a)2=4⇒(a+1a)2−(5)2=4⇒(a+1a)2−5=4⇒(a+1a)2=4+5⇒(a+1a)2=9⇒(a+1a)=9⇒(a+1a)=±3\Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - \Big(a - \dfrac{1}{a}\Big)^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - (\sqrt{5})^2 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 - 5 = 4 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 4 + 5 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big)^2 = 9 \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \sqrt{9} \\[1em] \Rightarrow \Big(a + \dfrac{1}{a}\Big) = \pm 3

Hence, (a+1a)=±3.\Big(a + \dfrac{1}{a}\Big) = \pm 3.

(ii) Given,

(a+1a)=±3.\Big(a + \dfrac{1}{a}\Big) = \pm 3.

Case 1:

(a+1a)=+3.\Big(a + \dfrac{1}{a}\Big) = +3.

We know that,

⇒(a+1a)3=a3+1a3+3(a+1a)\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)

Substituting values we get :

⇒33=a3+1a3+3×3⇒27=a3+1a3+9⇒a3+1a3=27−9=18.\Rightarrow 3^3 = a^3 + \dfrac{1}{a^3} + 3 \times 3 \\[1em] \Rightarrow 27 = a^3 + \dfrac{1}{a^3} + 9 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = 27 - 9 = 18.

Case 2:

(a+1a)=−3.\Big(a + \dfrac{1}{a}\Big) = -3.

We know that,

⇒(a+1a)3=a3+1a3+3(a+1a)\Rightarrow \Big(a + \dfrac{1}{a}\Big)^3 = a^3 + \dfrac{1}{a^3} + 3\Big(a + \dfrac{1}{a}\Big)

Substituting values we get :

⇒(−3)3=a3+1a3+3×(−3)⇒−27=a3+1a3−9⇒a3+1a3=−27+9=−18.\Rightarrow (-3)^3 = a^3 + \dfrac{1}{a^3} + 3 \times (-3) \\[1em] \Rightarrow -27 = a^3 + \dfrac{1}{a^3} - 9 \\[1em] \Rightarrow a^3 + \dfrac{1}{a^3} = -27 + 9 = -18.

Hence, a3+1a3=±18.a^3 + \dfrac{1}{a^3} = \pm 18.

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