Question

Question :-
If (a + [1]/[a])² = 3 , then a³⁹ + a²¹ - a²⁷ - a⁹ + 1 = ? (a>0)
​

Answer 1

Given,

$\longrightarrow\left(a+\dfrac{1}{a}\right)^2=3$

Taking square root,

$\longrightarrow a+\dfrac{1}{a}=\sqrt3$

$\longrightarrow\dfrac{a^2+1}{a}=\sqrt3$

$\longrightarrow a^2+1=a\sqrt3$

$\longrightarrow a^2-a\sqrt3+1=0$

Finding $a,$

$\longrightarrow a=\dfrac{\sqrt3\pm\sqrt{3-4}}{2}$

$\longrightarrow a=\dfrac{\sqrt3\pm i}{2}$

$\longrightarrow a=\dfrac{\sqrt3}{2}\pm\dfrac{1}{2}\,i$

Consider,

$\longrightarrow a=\dfrac{\sqrt3}{2}+\dfrac{1}{2}\,i$

Since $\cos\left(\dfrac{\pi}{6}\right)=\dfrac{\sqrt3}{2}$ and $\sin\left(\dfrac{\pi}{6}\right)=\dfrac{1}{2},$

$\longrightarrow a=\cos\left(\dfrac{\pi}{6}\right)+i\sin\left(\dfrac{\pi}{6}\right)$

By Euler's notation,

$\longrightarrow a=e^{\frac{i\pi}{6}}$

Now,

$\longrightarrow S=a^{39}+a^{21}-a^{27}-a^9+1$

$\longrightarrow S=e^{\frac{39i\pi}{6}}+e^{\frac{21i\pi}{6}}-e^{\frac{27i\pi}{6}}-e^{\frac{9i\pi}{6}}+1$

$\longrightarrow S=e^{\frac{13i\pi}{2}}+e^{\frac{7i\pi}{2}}-e^{\frac{9i\pi}{2}}-e^{\frac{3i\pi}{2}}+1$

Since $e^{\frac{i\pi}{2}}=i,$

$\longrightarrow S=i^{13}+i^7-i^9-i^3+1$

$\longrightarrow S=i-i-i+i+1$

$\longrightarrow\underline{\underline{S=1}}$

Consider,

$\longrightarrow a=\dfrac{\sqrt3}{2}-\dfrac{1}{2}\,i$

Since $\cos\left(-\dfrac{\pi}{6}\right)=\dfrac{\sqrt3}{2}$ and $\sin\left(-\dfrac{\pi}{6}\right)=-\dfrac{1}{2},$

$\longrightarrow a=\cos\left(-\dfrac{\pi}{6}\right)+i\sin\left(-\dfrac{\pi}{6}\right)$

By Euler's notation,

$\longrightarrow a=e^{-\frac{i\pi}{6}}$

Now,

$\longrightarrow S=a^{39}+a^{21}-a^{27}-a^9+1$

$\longrightarrow S=e^{-\frac{39i\pi}{6}}+e^{-\frac{21i\pi}{6}}-e^{-\frac{27i\pi}{6}}-e^{-\frac{9i\pi}{6}}+1$

$\longrightarrow S=e^{-\frac{13i\pi}{2}}+e^{-\frac{7i\pi}{2}}-e^{-\frac{9i\pi}{2}}-e^{-\frac{3i\pi}{2}}+1$

Since $e^{-\frac{i\pi}{2}}=-i=i^3,$

$\longrightarrow S=i^{39}+i^{21}-i^{27}-i^9+1$

$\longrightarrow S=-i+i+i-i+1$

$\longrightarrow\underline{\underline{S=1}}$

Hence (2) is the answer.

Answer 2

It's well known that

AM > or = GM

Here,

AM =Arithmetic Mean

GM=Geometric Mean

So

( a+1/a )/2 > or = (+ve or -ve)square root of (a×1/a)

=> (a+1/a) > or = (2) or (-2)

For the minimum value of (a+1/a), take a=1or (-1).

According to the question

a+1/a =-2

=> a =(-1)

So

Putting a =(-1) in

a⁴+a³+a²+a+1 we get

a⁴+a³+a²+a+1=1 as answer to this question.

I hope it helps