Answer:
(a) 
(b) Proved
Step-by-step explanation:
Given
--- the root
Solving (a): The polynomial
A cubic function is represented as:

Expand

Rewrite as:

The root is represented as:

By comparison:
![a = $\sqrt[3]{2}](https://tex.z-dn.net/?f=a%20%3D%20%24%5Csqrt%5B3%5D%7B2%7D)
![b = \sqrt[3]{4}$](https://tex.z-dn.net/?f=b%20%3D%20%5Csqrt%5B3%5D%7B4%7D%24)
So, we have:
![f = ($\sqrt[3]{2})^3 + 3*$\sqrt[3]{2}*\sqrt[3]{4}$*($\sqrt[3]{2} + \sqrt[3]{4}$) + (\sqrt[3]{4}$)^3](https://tex.z-dn.net/?f=f%20%3D%20%28%24%5Csqrt%5B3%5D%7B2%7D%29%5E3%20%2B%203%2A%24%5Csqrt%5B3%5D%7B2%7D%2A%5Csqrt%5B3%5D%7B4%7D%24%2A%28%24%5Csqrt%5B3%5D%7B2%7D%20%2B%20%5Csqrt%5B3%5D%7B4%7D%24%29%20%2B%20%28%5Csqrt%5B3%5D%7B4%7D%24%29%5E3)
Expand
![f = 2 + 3*$\sqrt[3]{2*4}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4](https://tex.z-dn.net/?f=f%20%3D%202%20%2B%203%2A%24%5Csqrt%5B3%5D%7B2%2A4%7D%2A%28%24%5Csqrt%5B3%5D%7B2%7D%20%2B%20%5Csqrt%5B3%5D%7B4%7D%24%29%20%2B%204)
![f = 2 + 3*$\sqrt[3]{8}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4](https://tex.z-dn.net/?f=f%20%3D%202%20%2B%203%2A%24%5Csqrt%5B3%5D%7B8%7D%2A%28%24%5Csqrt%5B3%5D%7B2%7D%20%2B%20%5Csqrt%5B3%5D%7B4%7D%24%29%20%2B%204)
![f = 2 + 3*2*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4](https://tex.z-dn.net/?f=f%20%3D%202%20%2B%203%2A2%2A%28%24%5Csqrt%5B3%5D%7B2%7D%20%2B%20%5Csqrt%5B3%5D%7B4%7D%24%29%20%2B%204)
![f = 2 + 6($\sqrt[3]{2} + \sqrt[3]{4}$) + 4](https://tex.z-dn.net/?f=f%20%3D%202%20%2B%206%28%24%5Csqrt%5B3%5D%7B2%7D%20%2B%20%5Csqrt%5B3%5D%7B4%7D%24%29%20%2B%204)
Evaluate like terms
![f = 6 + 6($\sqrt[3]{2} + \sqrt[3]{4}$)](https://tex.z-dn.net/?f=f%20%3D%206%20%2B%206%28%24%5Csqrt%5B3%5D%7B2%7D%20%2B%20%5Csqrt%5B3%5D%7B4%7D%24%29)
Recall that: ![r = $\sqrt[3]{2} + \sqrt[3]{4}$](https://tex.z-dn.net/?f=r%20%3D%20%24%5Csqrt%5B3%5D%7B2%7D%20%2B%20%5Csqrt%5B3%5D%7B4%7D%24)
So, we have:

Equate to 0

Rewrite as:

Express as a cubic function

Hence, the cubic polynomial is:

Solving (b): Prove that r is irrational
The constant term of
is -6
The divisors of -6 are: -6,-3,-2,-1,1,2,3,6
Calculate f(x) for each of the above values to calculate the remainder when f(x) is divided by any of the above values








For r to be rational;
The divisors of -6 must divide f(x) without remainder
i.e. Any of the above values must equal 0
<em>Since none equals 0, then r is irrational</em>