Question

$\huge{\underline{\underline{\bf Question}}$

if $\sf \bigg(a + \dfrac{1}{a} \bigg)^{2} = 3$ then :-

$\sf a^{39} + a ^{21} - a ^{27} + 1 = ?$

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Step by step explanation ✅

Good luck !​

Answer 1

GiveN :-

• $\sf \bigg(a + \dfrac{1}{a} \bigg)^{2} = 3$

To FinD :-

• $\sf a^{39} + a ^{21} - a ^{27} + 1$

SolutioN :-

Here it's given that ,

$\sf \to\pink{\bigg(a + \dfrac{1}{a} \bigg)^{2} = 3}$

And we need to find the value of $\sf a^{39} + a ^{21} - a ^{27} + 1$ . We will take the given equation and simplify it .

$\sf: \implies\bigg(a + \dfrac{1}{a} \bigg)^{2} = 3\\\\\sf:\implies a^2+\dfrac{1}{a^2} +2(a)\bigg(\dfrac{1}{a}\bigg)=3 \\\\\sf:\implies a^2+\dfrac{1}{a^2}+2=3\\\\\sf:\implies a^2+\dfrac{1}{a^2}=1 \\\\\sf:\implies \dfrac{a^4+1}{a^2}=1\\\\\sf:\implies a^4+1=a^2\\\\\sf:\implies a^4-a^2+1=0\\\\\sf:\implies (a^2+1)(a^4-a^2-1)=(a^2+1)0\\\\\sf:\implies a^2(a^4-a^2+1)+1(a^4-a^2+1)=0 \\\\\sf:\implies a^6-a^4+a^2+a^4-a^2+1=0\\\\\sf:\implies a^6+1=0\\\\\sf:\implies \boxed{\pink{\sf a ^6= -1 }}$

Also a⁶ can be written as (a³)² . So ,

$\sf:\implies a^6=1\\\\\sf:\implies (a^3)^2 = -1 \\\\\sf:\implies a^3=\sqrt{-1} \\\\\sf:\implies \boxed{\pink{\sf{a^3} = i \:\:\;\:\sf{(iota)}}}$

$\rule{200}2$

Now we need to find out the value of a³⁹ + a²¹ - a²⁷ + 1 . Let the required answer be K . So ,

$\red{\bigstar}\underline{\textsf{ Put on the respective values :- }}$

$\sf:\implies K = a^{39}+a^{21}-a^{27}+1 \\\\\sf:\implies K = a^{21}( a^{18}+1-a^6) +1\\\\\sf:\implies K = a^{21}[ (a^6)^3 + 1 - a^6] +1\\\\\sf:\implies K = a^{21}[ (-1)^3+1-(-1)] +1\\\\\sf:\implies K = a^{21}[ -1+1+1] +1\\\\\sf:\implies K = a^{21}[ 2-1] + 1 \\\\\sf:\implies K = a^{21} + 1 \\\\\sf:\implies K = ( a^{18}.a^{3} ) +1 \\\\\sf:\implies K = [ (a^6)^3 \times a^3]+1 \\\\\sf:\implies K = [ (-1)^3. i + 1 ] \\\\\sf:\implies K = (-1)(i) +1 \\\\\sf:\implies\underset{\blue{\sf Required\ Answer }}{\underbrace{\boxed{\pink{\frak{ K = -i + 1 = -\sqrt{-1}+1}}}}}$