Question

(a) Determine a cubic polynomial with integer coefficients which has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

(b) Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.

Answer 1

(a)
`x\³ - 6x - 6`

(b) Proved

Explanation:

Given

`r = $\sqrt[3]{2} + \sqrt[3]{4}$` --- the root

Solving (a): The polynomial

A cubic function is represented as:

`f = (a + b)^3`

Expand

`f = a^3 + 3a^2b + 3ab^2 + b^3`

Rewrite as:

`f = a^3 + 3ab(a + b) + b^3`

The root is represented as:

`r=a+b`

By comparison:

`a = $\sqrt[3]{2}`

`b = \sqrt[3]{4}$`

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`

Expand

`f = 2 + 3*$\sqrt[3]{2*4}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4`

`f = 2 + 3*$\sqrt[3]{8}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4`

`f = 2 + 3*2*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4`

`f = 2 + 6($\sqrt[3]{2} + \sqrt[3]{4}$) + 4`

Evaluate like terms

`f = 6 + 6($\sqrt[3]{2} + \sqrt[3]{4}$)`

Recall that:
`r = $\sqrt[3]{2} + \sqrt[3]{4}$`

So, we have:

`f = 6 + 6r`

Equate to 0

`f - 6 - 6r = 0`

Rewrite as:

`f - 6r - 6 = 0`

Express as a cubic function

`x^3 - 6x - 6 = 0`

Hence, the cubic polynomial is:

`f(x) = x^3 - 6x - 6`

Solving (b): Prove that r is irrational

The constant term of
`x^3 - 6x - 6 = 0` 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

`f(-6) = (-6)^3 - 6*-6 - 6 = -186`

`f(-3) = (-3)^3 - 6*-3 - 6 = -15`

`f(-2) = (-2)^3 - 6*-2 - 6 = -2`

`f(-1) = (-1)^3 - 6*-1 - 6 = -1`

`f(1) = (1)^3 - 6*1 - 6 = -11`

`f(2) = (2)^3 - 6*2 - 6 = -10`

`f(3) = (3)^3 - 6*3 - 6 = 3`

`f(6) = (6)^3 - 6*6 - 6 = 174`

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

Since none equals 0, then r is irrational