# Thread: a^4+1/a^4=119. Then a^3+1/a^3

1. ## a^4+1/a^4=119. Then a^3+1/a^3

$a^4+\frac{1}{a^4}=119$Then find $a^3+\frac{1}{a^3}$

I know $a^3+b^3=(a+b)(a^2+b^2-ab)$ and $a^4+b^4+a^2b^2=(a^2+b^2+ab)(a^2+b^2-ab)$

I try combining both but end up with equation in terms of a to the power 8, 7, 6 etc.

2. ## Re: a^4+1/a^4=119. Then a^3+1/a^3

$a^4+\dfrac{1}{a^4}+1 = \left( a^2+\dfrac{1}{a^2}+1 \right)\left( a^2+\dfrac{1}{a^2} - 1 \right) = 119+1 = 120$

Let $x = a^2+\dfrac{1}{a^2}$. Then $(x+1)(x-1) = 120$. So, $x^2-1 = 120$ and $x^2 = 121$. This gives $x = \pm 11$.

Then $\pm 11 = a^2+\dfrac{1}{a^2}$, so $a^4\pm 11a^2+1=0$. Then $a^2 = \dfrac{11 \pm \sqrt{117} }{2}$ or $a^2 = \dfrac{ -11 \pm \sqrt{117} }{2}$. Since any real number squared must be positive, it must be the first one.

So, $a = \sqrt{ \dfrac{ 11\pm \sqrt{117} }{2} }$ or $a = -\sqrt{ \dfrac{ 11\pm \sqrt{117} }{2} }$. This gives:

\begin{align*}\left|a^3+\dfrac{1}{a^3}\right| & = \left( \dfrac{ 11\pm \sqrt{117} }{2} \right)^{3/2} + \left( \dfrac{ 11\pm \sqrt{117} }{2} \right)^{-3/2} \\ & = \left( \dfrac{ 11\pm 3\sqrt{13} }{2}\cdot \dfrac{2}{2} \right)^{3/2} + \left( \dfrac{2}{ 11\pm 3\sqrt{13} }\cdot \dfrac{ 11\mp 3\sqrt{13} }{ 11\mp 3\sqrt{13} } \right)^{3/2} \\ & = \left( \dfrac{ 22\pm 6\sqrt{13} }{4} \right)^{3/2} + \left( \dfrac{ 22 \mp 6\sqrt{13} }{ 4 } \right)^{3/2}\end{align*}

$22\pm 6\sqrt{13} = 9\pm 6\sqrt{13} + 13 = (\sqrt{13} \pm 3)^2$

Since the square root of a positive number is positive, I put the $\sqrt{13}$ first since $\sqrt{13}-3>0$ and $(\sqrt{13} \pm 3)^2 = (3\pm \sqrt{13})^2$.

Hence, we have

\begin{align*}\left( \dfrac{ 22\pm 6\sqrt{13} }{4} \right)^{3/2} + \left( \dfrac{ 22 \mp 6\sqrt{13} }{ 4 } \right)^{3/2} & = \dfrac{ \left[\left(\sqrt{13}\pm 3\right)^2\right]^{3/2} + \left[\left(\sqrt{13}\mp 3\right)^2\right]^{3/2} }{2^3} \\ & = \dfrac{ \left(13\sqrt{13} \pm 3\cdot 3\cdot 13 + 3\cdot 3^2\sqrt{13} \pm 3^3 \right) + \left(13\sqrt{13} \mp 3\cdot 3\cdot 13 + 3\cdot 3^2\sqrt{13} \mp 3^3\right) }{8} \\ & = \dfrac{80\sqrt{13}}{8} \\ & = 10\sqrt{13}\end{align*}

Hence, $a^3+\dfrac{1}{a^3} = \pm 10\sqrt{13}$.

3. ## Re: a^4+1/a^4=119. Then a^3+1/a^3

Originally Posted by SlipEternal
$a^4+\dfrac{1}{a^4}+1 = \left( a^2+\dfrac{1}{a^2}+1 \right)\left( a^2+\dfrac{1}{a^2} - 1 \right) = 119+1 = 120$

Let $x = a^2+\dfrac{1}{a^2}$. Then $(x+1)(x-1) = 120$. So, $x^2-1 = 120$ and $x^2 = 121$. This gives $x = \pm 11$.

Then $\pm 11 = a^2+\dfrac{1}{a^2}$, so $a^4\pm 11a^2+1=0$. Then $a^2 = \dfrac{11 \pm \sqrt{117} }{2}$ or $a^2 = \dfrac{ -11 \pm \sqrt{117} }{2}$. Since any real number squared must be positive, it must be the first one.

So, $a = \sqrt{ \dfrac{ 11\pm \sqrt{117} }{2} }$ or $a = -\sqrt{ \dfrac{ 11\pm \sqrt{117} }{2} }$. This gives:

\begin{align*}\left|a^3+\dfrac{1}{a^3}\right| & = \left( \dfrac{ 11\pm \sqrt{117} }{2} \right)^{3/2} + \left( \dfrac{ 11\pm \sqrt{117} }{2} \right)^{-3/2} \\ & = \left( \dfrac{ 11\pm 3\sqrt{13} }{2}\cdot \dfrac{2}{2} \right)^{3/2} + \left( \dfrac{2}{ 11\pm 3\sqrt{13} }\cdot \dfrac{ 11\mp 3\sqrt{13} }{ 11\mp 3\sqrt{13} } \right)^{3/2} \\ & = \left( \dfrac{ 22\pm 6\sqrt{13} }{4} \right)^{3/2} + \left( \dfrac{ 22 \mp 6\sqrt{13} }{ 4 } \right)^{3/2}\end{align*}

$22\pm 6\sqrt{13} = 9\pm 6\sqrt{13} + 13 = (\sqrt{13} \pm 3)^2$

Since the square root of a positive number is positive, I put the $\sqrt{13}$ first since $\sqrt{13}-3>0$ and $(\sqrt{13} \pm 3)^2 = (3\pm \sqrt{13})^2$.

Hence, we have

\begin{align*}\left( \dfrac{ 22\pm 6\sqrt{13} }{4} \right)^{3/2} + \left( \dfrac{ 22 \mp 6\sqrt{13} }{ 4 } \right)^{3/2} & = \dfrac{ \left[\left(\sqrt{13}\pm 3\right)^2\right]^{3/2} + \left[\left(\sqrt{13}\mp 3\right)^2\right]^{3/2} }{2^3} \\ & = \dfrac{ \left(13\sqrt{13} \pm 3\cdot 3\cdot 13 + 3\cdot 3^2\sqrt{13} \pm 3^3 \right) + \left(13\sqrt{13} \mp 3\cdot 3\cdot 13 + 3\cdot 3^2\sqrt{13} \mp 3^3\right) }{8} \\ & = \dfrac{80\sqrt{13}}{8} \\ & = 10\sqrt{13}\end{align*}

Hence, $a^3+\dfrac{1}{a^3} = \pm 10\sqrt{13}$.
THanks and very very sorry. I asked my friend in phone. He said that we should find a^3-1/a^3 instead of a^3+1/a^3.

4. ## Re: a^4+1/a^4=119. Then a^3+1/a^3

$a^3-b^3 = (a-b)(a^2+b^2+ab)$

So, we have

\begin{align*}\left| a-\dfrac{1}{a} \right| & = \sqrt{ \dfrac{ 11\pm 3\sqrt{13} }{2} } - \sqrt{ \dfrac{2}{ 11 \pm 3\sqrt{13} } } \\ & = \dfrac{ \sqrt{13} \pm 3 }{2} - \dfrac{ \sqrt{13} \mp 3 }{2} \\ & = \pm 3\end{align*}

Above, I showed $x = a^2+\dfrac{1}{a^2} = 11$.

So, $a-\dfrac{1}{a} = \pm 3$, and $a^3 - \dfrac{1}{a^3} = \left( a - \dfrac{1}{a} \right) \left( a^2 + \dfrac{1}{a^2} + 1\right) = (\pm 3)(11+1) = \pm 36$

5. ## Re: a^4+1/a^4=119. Then a^3+1/a^3

Thank you very much.

I don't understand how the first thing is same as the second line?

Originally Posted by SlipEternal
\begin{align*} \sqrt{ \dfrac{ 11\pm 3\sqrt{13} }{2} } - \sqrt{ \dfrac{2}{ 11 \pm 3\sqrt{13} } } & = \dfrac{ \sqrt{13} \pm 3 }{2} - \dfrac{ \sqrt{13} \mp 3 }{2}\end{align*}
[/tex]

6. ## Re: a^4+1/a^4=119. Then a^3+1/a^3

I did all of the work for that in the post above.

7. ## Re: a^4+1/a^4=119. Then a^3+1/a^3

Thank you. I did not go through your first post properly.

8. ## Re: a^4+1/a^4=119. Then a^3+1/a^3

Here's a way without actually solving for $a$. First consider

$\left(a^2 + \dfrac{1}{a^2} \right)^2 = a^4 + 2 + \dfrac{1}{a^4} = 119 + 2 = 121$

so $a^2 + \dfrac{1}{a^2}=11$ (already established)!

Then consider

$\left(a + \dfrac{1}{a} \right)^4 = a^4 + 4a^2 + 6 + \dfrac{4}{a^2} + \dfrac{1}{a^4} = 119 + 4 \cdot 11 + 6 = 169$

so

$a + \dfrac{1}{a} = \pm \sqrt{13}$

Then consider

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

from which we obtain

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

Then substitute what we know giving

$a^3 + \dfrac{1}{a^3} = \pm 13\sqrt{13} - 3(\pm\sqrt{13}) = \pm 10\sqrt{13}$.

9. ## Re: a^4+1/a^4=119. Then a^3+1/a^3

There's a typo that I can't fix. The $3a^2$ should be a $3a$.