MHB Can you simplify this surd expression without a calculator?

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The discussion focuses on simplifying the surd expression without a calculator. Participants explore the roots of the expressions involved, particularly using the forms of polynomials and recursive sequences to derive values for a and b. The simplification leads to the conclusion that the expression equals 1. Key techniques include equating rational and irrational parts and leveraging properties of powers of surds. Ultimately, the final result of the simplification is confirmed as 1.
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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)
 
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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)
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
 
@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
 
topsquark said:
@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)
 
(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
 
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:

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