# Math Help - prove that

1. ## prove that

prove that

2. Hello, perash!

I got the first half . . .

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

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

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

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

3. 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?

4. 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...

5. What do you mean? It did cancel out in part A or did I misunderstand your question?

Lol I was too slow.