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