Question 1174265
.
(a) Find a cubic polynomial with integer coefficients that has
(cube root of 2) + (cube root of 4) as a root.
(b) Prove that the (cube root of 2) + (the cube root of 4) is irrational.
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<U>Part (a)</U>


<pre>
Let r = {{{root(3,2)}}} + {{{root(3,4)}}}.


Then,  since  (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + 3ab*(a+b) + b^3


    r^3 = {{{(root(3,2))^3}}} + {{{3*root(3,2)*root(3,4)}}}.{{{(root(3,2) + root(3,4))}}} + {{{(root(3,4))^3}}} = 

        = 2 + {{{3*root(3,8)}}}.{{{(root(3,2)+root(3,4))}}} + 4 = 2 + 3*2*r + 4 = 6 + 6r.


It means that  r = {{{root(3,2)}}} + {{{root(3,4)}}}  is the root to this equation


    r^3 - 6r - 6 = 0.      (*)


<U>ANSWER</U>.   {{{root(3,2)}}} + {{{root(3,4)}}}  is the root of the cubic polynomial  x^3 - 6x - 6.
</pre>


Part (a) is solved.




<U>Part (b)</U>


<pre>
In part (a), I proved that the number  r = {{{root(3,2)}}} + {{{root(3,4)}}}  is the root of the cubic equation


    x^3 - 6x - 6 = 0.     (**)


Therefore, due to the Rational root theorem, if the number "r" is rational, it must divide the constant term of 6, 
i.e. r must be one of the numbers +/-1, +/-2, +/-3, +/-6.


But it is easy to check that no one of these divisors of 6 IS NOT THE ROOT to equation (**).


Indeed, for these values of x, the values of the polynomial  f(x) = x^3 - 6x -6  are given in the Table


    x       -1      1     -2      2     -3      3     -6     6

    f(x)    -1    -11     -2    -10    -15      3    -186   174


and no one of these values of the polynomial  f(x)  is equal to  0  (zero).
</pre>

Part (b) is solved, too.