Find the value of \(\frac{a^2}{a^4+a^2+1}\) if \(\frac{a}{a^2+a+1}=\frac{1}{6}\).
For some reason, I keep getting 1/36 every time I attempt it, and the answer is 1/24.
Please, someone give me a helping hint
The hint is that \((a^2+a+1)^2 \ne a^4+a^2+1\)
\(\frac{a}{a^2+a+1}=\frac{1}{6}\\ (\frac{a}{a^2+a+1})^2=\frac{1}{36}\\ (\frac{a}{a^2+a+1})^2=\frac{1}{36}\\ (\frac{a^2+a+1}{a})^2=36\\ \frac{a^4+2a^3+3a^2+2a+1}{a^2}=36\\ \frac{a^4+a^2+1+2a^3+2a^2+2a}{a^2}=36\\ \frac{(a^4+a^2+1)+2a(a^2+a+1)}{a^2}=36\\ \frac{(a^4+a^2+1)}{a^2}+\frac{2a(a^2+a+1)}{a^2}=36\\ \frac{(a^4+a^2+1)}{a^2}+2*\frac{(a^2+a+1)}{a}=36\\ \frac{(a^4+a^2+1)}{a^2}+12=36\\ \frac{(a^4+a^2+1)}{a^2}=24\\ \frac{a^2}{(a^4+a^2+1)}=\frac{1}{24}\\ \)
You can factor it as follows:
a^2 - 5 a + 1 = -1/4 (-2 a + sqrt(21) + 5) * (2 a + sqrt(21) - 5)
Split into 2 equations:
1/4 (-2 a + sqrt(21) + 5) =0
a =4.7912878....
(2 a + sqrt(21) - 5) =0
a =0.20871215.......
Sub any of the 2 values of a into the first part and you get:
= 1 / 24
The hint is that \((a^2+a+1)^2 \ne a^4+a^2+1\)
\(\frac{a}{a^2+a+1}=\frac{1}{6}\\ (\frac{a}{a^2+a+1})^2=\frac{1}{36}\\ (\frac{a}{a^2+a+1})^2=\frac{1}{36}\\ (\frac{a^2+a+1}{a})^2=36\\ \frac{a^4+2a^3+3a^2+2a+1}{a^2}=36\\ \frac{a^4+a^2+1+2a^3+2a^2+2a}{a^2}=36\\ \frac{(a^4+a^2+1)+2a(a^2+a+1)}{a^2}=36\\ \frac{(a^4+a^2+1)}{a^2}+\frac{2a(a^2+a+1)}{a^2}=36\\ \frac{(a^4+a^2+1)}{a^2}+2*\frac{(a^2+a+1)}{a}=36\\ \frac{(a^4+a^2+1)}{a^2}+12=36\\ \frac{(a^4+a^2+1)}{a^2}=24\\ \frac{a^2}{(a^4+a^2+1)}=\frac{1}{24}\\ \)