(a+1/a)^2=3 Then A^30+a^24+a^18+a^12+a^6+1

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Jul 03, 2024

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Solving the Equation: $\left(a+\frac{1}{a}\right)^2=3$ and Finding the Value of $a^{30}+a^{24}+a^{18}+a^{12}+a^6+1$

The Given Equation:

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

Simplifying the Equation:

$a^2+2a\cdot\frac{1}{a}+\left(\frac{1}{a}\right)^2=3$

$a^2+2+\frac{1}{a^2}=3$

$a^2+\frac{1}{a^2}-1=0$

Let's Introduce a New Variable:

Let $x=a^2+\frac{1}{a^2}$. Then, the equation becomes:

$x-1=0$

$x=1$

Now, We Can Express $a^2$ and $\frac{1}{a^2}$ in Terms of $x$:

$a^2+\frac{1}{a^2}=1$

$a^2=1-\frac{1}{a^2}$

Finding the Value of $a$:

$a^2=1-\frac{1}{a^2}$

$a^4-1=0$

$(a^2-1)(a^2+1)=0$

$a^2=1\quad\text{or}\quad a^2=-1$

$a=1,-1,\quad i,-i$

Now, We Can Calculate the Value of $a^{30}+a^{24}+a^{18}+a^{12}+a^6+1$:

We have four possible values of $a$, so we will calculate the value of the expression for each of them:

Case 1: $a=1$

$a^{30}+a^{24}+a^{18}+a^{12}+a^6+1=1+1+1+1+1+1=6$

Case 2: $a=-1$

$a^{30}+a^{24}+a^{18}+a^{12}+a^6+1=(-1)^{30}+(-1)^{24}+(-1)^{18}+(-1)^{12}+(-1)^6+1=6$

Case 3: $a=i$

$a^{30}+a^{24}+a^{18}+a^{12}+a^6+1=i^{30}+i^{24}+i^{18}+i^{12}+i^6+1=6$

Case 4: $a=-i$

$a^{30}+a^{24}+a^{18}+a^{12}+a^6+1=(-i)^{30}+(-i)^{24}+(-i)^{18}+(-i)^{12}+(-i)^6+1=6$

Conclusion:

The value of $a^{30}+a^{24}+a^{18}+a^{12}+a^6+1$ is $\boxed{6}$ for all possible values of $a$ that satisfy the equation $\left(a+\frac{1}{a}\right)^2=3$.