# Find $\frac{a^3}{a^6 + 1}$ given $a$ is a root of a quadratic equation

If $a$ is a root of the equation $x^2 - 3x + 1 = 0$, then find the value of $\frac{a^3}{a^6 + 1}$.

So, I figured we can use the quadratic formula, and formed the following equation: $$a=\frac{-(-3)+\sqrt{9-4}}{2(1)}\implies a=\frac{3+\sqrt5}2$$ But what I am thinking is, if I begin to find the required value, it will take me hours. And I believe, there must be some shortcut to this question. I had tried to solve this question with the manual process but it took me a lot of squares (one was $2012^2$!)

Can someone please help me.

Thank you.

• $a^2-3a+1 = 0 \iff a+a^{-1} = 3$ and $\frac{a^3}{a^6+1} = \frac{1}{a^3+a^{-3}}$. – achille hui Feb 2 '14 at 9:13

## 4 Answers

So $$a^2 + 1 = 3a$$ and this gives: $$\frac{a}{a^2 + 1} = \frac{1}{3},$$ and $$(a^2 + 1)^2 = 9a^2 \implies a^4 + 1 = 7a^2.$$ So $$\frac{a^2}{a^4 - a^2 + 1} = \frac{a^2}{6a^2} = \frac{1}{6}.$$ And finally $$\frac{a^3}{a^6 + 1} = \frac{a}{a^2 +1}\cdot \frac{a^2}{a^4 - a^2 + 1} = \frac{1}{3}\cdot \frac{1}{6} = \frac{1}{18}.$$

Since $a$ is a root of the quadratic, you know $a^2 -3a +1 = 0$, hence, $a^2 = 3a-1$.

This means any power of $a$ can be expressed as a linear combination of $a$ and $1$ in a natural fashion.

E.g. $a^3 = a(3a-1) = 3a^2 -a = 3(3a-1) -a = 8a -3$.

$a^2=3a-1$ then $$\frac{a^3}{a^6 + 1} \\=\frac{a^3}{(a^2+1)(a^4-a^2 + 1)} \\=\frac{a^3}{(3a)(9a^2-6a+1-3a+1+1)} \\=\frac{a^2}{(3)(9a^2-9a+3)} \\=\frac{a^2}{(9)(3a^2-3a+1)} \\=\frac{a^2}{(9)(9a-3-3a+1)} \\=\frac{a^2}{(9)(6a-2)} \\=\frac{3a-1}{(18)(3a-1)} \\=\frac{1}{18}$$

• Your method is flawed since $a^2+1\neq3a$. – user93957 Feb 2 '14 at 9:25
• @Aðøbe, no, this is true, it follows directly from the quadratic equation!! – Joachim Feb 2 '14 at 9:32
• @Aðøbe a is aroot – Semsem Feb 2 '14 at 9:34
• @Semsem Sorry, it's because in your first version you wrote $a^2=3a\color{red}+1$ so I naturally concluded from it that $a^2+1\neq 3a$. – user93957 Feb 2 '14 at 9:38
• @Semsem In your fourth equation, your numerator is $a$, shouldn't it be $a^2$ ? – Gaurang Tandon Feb 2 '14 at 10:00

As $a\ne0,$ we have $\displaystyle a^2-3a+1=0\iff a^2+1=3a\implies a+\frac1a=3$

and $\displaystyle a^3+\dfrac1{a^3}=\left(a+\frac1a\right)^3-3\left(a+\frac1a\right)^3$

$$\text{Again, }\frac{a^3}{a^6+1}=\frac1{a^3+\dfrac1{a^3}}$$

• @GaurangTandon, I thought of adding a relatively simpler derivation. Try to solve similar problems – lab bhattacharjee Feb 2 '14 at 13:00