Can you simplify this surd expression without a calculator?

[MarkFL]

Without the use of a calculator, and showing your work, simplify:

$$\frac{1}{2}\left(\left(239+169\sqrt{2}\right)^{ \frac{1}{7}}-\left(29\sqrt{2}-41\right)^{ \frac{1}{5}}\right)$$

edit: My apologies...I was careless in my first statement of the problem...(Nod)

 

[soroban]

Hello, MarkFL!

Without the use of a calculator, and showing your work, simplify:

. . \tfrac{1}{2}\left(\left(239+169\sqrt{2}\right)^{ \frac{1}{7}}-\left(29\sqrt{2}-41\right)^{\frac{1}{5}}\right)

Spoiler

We find that: .239 + 169\sqrt{2} \:=\: (1+\sqrt{2})^7

. . .and that: .. 29\sqrt{2} - 41 \:=\: (\sqrt{2} - 1)^5

So we have: .\tfrac{1}{2}\left(\left[(1+\sqrt{2})^7\right]^{\frac{1}{7}} - \left[(\sqrt{2}-1)^5\right]^{\frac{1}{5}}\right)

. . . . . . . =\;\tfrac{1}{2}\bigg(\left[1 + \sqrt{2}\right] - \left[\sqrt{2}-1\right]\bigg)

. . . . . . . =\;\tfrac{1}{2}\big(1 + \sqrt{2} - \sqrt{2} + 1\big)

. . . . . . . =\;\tfrac{1}{2}(2)

. . . . . . . =\;1

 

[topsquark]

@Soroban. You are the master of coming up with solutions that seem to come out of the blue. How the heck did you get the first two lines of your solution?

-Dan

 

[Nono713]

@Soroban. You are the master of coming up with solutions that seem to come out of the blue. How the heck did you get the first two lines of your solution?

-Dan

My thoughts exactly. I was sitting there, thinking, "okay". You must teach me! (Rofl)

 

[kaliprasad]

(239+169√2)^1/7

the root can be of the form (a+b √2) and expanding and equating the rational parts of both sides and irrational parts of both sides we get 2 equations in a and b
then solving them we get a = b = 1

I know these are polynomials of degree 7 and 6 and solving is not simple so assuming a and b as integers
we can put the values on both sides and get a and b.

simlilarly for we can find 5th root of 2nd expression

 

[MarkFL]

This is the observation I made before constructing this problem:

$$\left(\sqrt{2}+1 \right)^0=0\cdot\sqrt{2}+1$$

$$\left(\sqrt{2}+1 \right)^1=1\cdot\sqrt{2}+1$$

$$\left(\sqrt{2}+1 \right)^2=2\cdot\sqrt{2}+3$$

$$\left(\sqrt{2}+1 \right)^3=5\cdot\sqrt{2}+7$$

$$\left(\sqrt{2}+1 \right)^4=12\cdot\sqrt{2}+17$$

Now, we may generalize to say:

$$\left(\sqrt{2}+1 \right)^n=U_n\sqrt{2}+V_n$$

We may further generalize and write:

$$\left(\sqrt{2}-1 \right)^n=(-1)^{n-1}\left(U_n\sqrt{2}-V_n \right)$$

The parameters may be defined recursively as:

$$U_{n+1}=2U_{n}+U_{n-1}$$ where $$U_0=0,\,U_1=1$$

$$V_{n+1}=2V_{n}+V_{n-1}$$ where $$V_0=1,\,V_1=1$$

We then find:

$$U_5=29,\,U_7=169$$

$$V_5=41,\,V_7=239$$

and so:

$$29\sqrt{2}-41=(-1)^4\left(\sqrt{2}-1 \right)^5\,\therefore\,\left(29\sqrt{2}-41 \right)^{\frac{1}{5}}=\sqrt{2}-1$$

$$169\sqrt{2}+239=\left(\sqrt{2}+1 \right)^7\,\therefore\,\left(169\sqrt{2}+239 \right)^{\frac{1}{7}}=\sqrt{2}+1$$

Hence:

$$\frac{1}{2}\left(\left(169\sqrt{2}+239 \right)^{\frac{1}{7}}-\left(29\sqrt{2}-41 \right)^{\frac{1}{5}} \right)=\frac{1}{2}\left(\sqrt{2}+1-\sqrt{2}+1 \right)=\frac{1}{2}\cdot2=1$$