Question

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

Answer 1

Answer:

(a) $x\³ - 6x - 6$

(b) Proved

Step-by-step 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