• Jun 16th 2012, 07:21 AM
Kaloda
• Jun 16th 2012, 08:50 AM
simamura
We will need two formulas:
$x^3-y^3=(x-y)(x^2+xy+y^2)$
and
$(x-y)^2=x^2-2xy+y^2$

Let $\sqrt[12]x=a$

Then we are given that $a^3- \frac{1}{a^3}=14$

We need to show that $a^2+\frac{1}{a^2}=6$.

Expand what we are given by first formula: $a^3- \frac{1}{a^3}=\left(a-\frac 1a\right)\left(a^2+\frac aa+\frac {1}{a^2}\right)=14$
or
$\left(a-\frac 1a\right)\left(a^2+\frac {1}{a^2}+1\right)=14$

Since $\left(a-\frac1a\right)^2=a^2-2 \frac aa+\frac{1}{a^2}=a^2+\frac{1}{a^2}-2$ then $\left(a-\frac1a\right)^2+2=a^2+\frac{1}{a^2}$
Now, equation can be rewritten as
$\left(a-\frac 1a\right)\left(\left(a-\frac1a\right)^2+2+1\right)=14$
Or
$\left(a-\frac 1a\right)\left(\left(a-\frac1a\right)^2+3\right)=14$
For simplicity, make substitution $t=a-\frac 1a$

Then we have that $t(t^2+3)=14$
There is only one real solution t=2, so $a-\frac 1a=2$
Finally, $a^2+\frac{1}{a^2}=\left(a-\frac1a\right)^2+2=2^2+2=6$
That's all.
• Jun 16th 2012, 10:09 AM
richard1234
You're given

Let $\sqrt[6]{x} + \frac{1}{\sqrt[6]{x}} = k$. Cubing both sides of this equation yields

$\sqrt[6]{x}^3 + 3 \sqrt[6]{x}^2(\frac{1}{\sqrt[6]{x}}) + 3\sqrt[6]{x}(\frac{1}{\sqrt[6]{x}})^2 + (\frac{1}{\sqrt[6]{x}})^3 = k^3$

$\sqrt{x} + \frac{1}{\sqrt{x}} + 3(\sqrt[6]{x} + \frac{1}{\sqrt[6]{x}}) = k^3$, substitute $\sqrt[6]{x} + \frac{1}{\sqrt[6]{x}}$ with $k$.

$\sqrt{x} + \frac{1}{\sqrt{x}} + 3k = k^3$. Here, it helps to know that $\sqrt{x} + \frac{1}{\sqrt{x}}$ is. However, we know that $\sqrt[4]{x} - \frac{1}{\sqrt[4]{x}} = 14$. Squaring both sides of this equation yields

$\sqrt{x} + \frac{1}{\sqrt{x}} - 2 = 14^2 = 196 \Rightarrow \sqrt{x} + \frac{1}{\sqrt{x}} = 198$. Substitute into the previous equation to obtain

$198 + 3k = k^3$. Here it is evident that $k = 6$ is a root (the other two roots are non-real). Hence we are done.
• Jun 16th 2012, 10:30 AM
Kaloda