prove that

• Sep 6th 2008, 04:17 PM
perash
prove that
• Sep 6th 2008, 09:23 PM
Soroban
Hello, perash!

I got the first half . . .

Quote:

$\text{Let: }\:A \;=\;\sqrt[3]{3 + 12\sqrt[3]{11} - 6\sqrt[3]{121}}\;\text{ and }\;B \;=\;\sqrt[3]{16 - 27\sqrt[3]{11} + 9\sqrt[3]{121}}$

$\text{Show (without using a calculator) that:}$

. . $(a)\;\;A + B \:=\:1\qquad\qquad (b)\;\;A\cdot B \:< \:\frac{1}{4}$

I happened to notice that: . $\begin{array}{ccc}3 + 12\sqrt[3]{11} - 6\sqrt[3]{121} &=& \left(\sqrt[3]{11} - 2\right)^3 \\ \\[-3mm] 16 -27\sqrt[3]{11} + 9\sqrt[3]{121} &=& \left(3 - \sqrt[3]{11}\right)^3 \end{array}$

(a) Therefore: . $A + B \;=\;\sqrt[3]{\left(\sqrt[3]{11}-2\right)^3} + \sqrt[3]{\left(3-\sqrt[3]{11}\right)^3} \;=\;\left(\sqrt[3]{11} - 2\right) + \left(3 - \sqrt[3]{11}\right) \;=\;1$

• Sep 7th 2008, 06:06 PM
sportsfan93b
Quote:

Originally Posted by Soroban
Hello, perash!

I got the first half . . .

I happened to notice that: . $\begin{array}{ccc}3 + 12\sqrt[3]{11} - 6\sqrt[3]{121} &=& \left(\sqrt[3]{11} - 2\right)^3 \\ \\[-3mm] 16 -27\sqrt[3]{11} + 9\sqrt[3]{121} &=& \left(3 - \sqrt[3]{11}\right)^3 \end{array}$

(a) Therefore: . $A + B \;=\;\sqrt[3]{\left(\sqrt[3]{11}-2\right)^3} + \sqrt[3]{\left(3-\sqrt[3]{11}\right)^3} \;=\;\left(\sqrt[3]{11} - 2\right) + \left(3 - \sqrt[3]{11}\right) \;=\;1$

In your final solution wouldnt the cube route and the 3rd pwr cancel out?
• Sep 7th 2008, 06:14 PM
Jhevon
Quote:

Originally Posted by sportsfan93b
In your final solution wouldnt the cube root and the 3rd pwr cancel out?

what do you mean? they did cancel out...
• Sep 7th 2008, 06:14 PM
11rdc11
What do you mean? It did cancel out in part A or did I misunderstand your question?

Lol I was too slow.