Set . Then . 

 can be rewritten as

Substituting  for  and  for , the equation becomes 

a quadratic equation in .

This can be solved using the  method. We are looking for two integers whose sum is  and whose product is . Through some trial and error, the integers are found to be  and , so the above equation can be rewritten, and solved using grouping, as

By the Zero Product Principle, one of these factors is equal to zero:

Either: 

Substituting back:

, or 

Or:

Substituting back:

, or 

Define . Then, if , it follows that .

By the Intermediate Value Theorem (IVT), if  is a continuous function, and  and  are of unlike sign, then  for some .  and  are both continuous everywhere, so  is a continuous function, so the IVT applies here.

Evaluate  for each of the following values: :

Only in the case of  does it hold that  assumes a different sign at the endpoints - . By the IVT, , and , for some .

Let  be a cube root of . The question is to find a solution of the equation

.

One way to solve this is to add 64 to both sides:

64 is a perfect cube, so, as the sum of cubes, the left expression can be factored:

We can set both factors equal to zero and solve:

 is a cube root of ; however, this is not one of the choices.

Setting 

,

we can make use of the quadratic formula, setting  in the following:

 and  are both cube roots of ;  is not a choice, but  is. 

A fourth-degree, or quartic, polynomial has four zeroes, if a zero of multiplicity  is counted  times. Since its lead term is , we know that

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since  is such a polynomial, then, since  is a zero, so is its complex conjugate ; similarly, since  is a zero, so is its complex conjugate . Substituting these four values for the four  values: 

This can be rewritten as

or 

Multiply the first two factors using the difference of squares pattern, then the square of a binomial pattern:

Multiply the last two factors similarly:

Thus,

Multiply:

________________

.

Define . Then, if , it follows that .

By the Intermediate Value Theorem (IVT), if  is a continuous function, and  and  are of unlike sign, then  for some . As a polynomial,  is a continuous function, so the IVT applies here. 

Evaluate  for each of the following values: :

Only in the case of  does it hold that  assumes different signs at the endpoints - . By the IVT, , and , for some .

Factor the trinomial.

The multiples of the first term is .

The multiples of the third term is .

We can then factor using these terms.

Set the equation to zero.

This means that each product will equal zero.

The roots are either 

The answer is:  

From the discriminant, , we know that this equation will have two solutions:

Next, factor the equation .

Finally, solve for .

One way to work this problem is as follows:

Factor  using the difference of squares pattern:

Consequently, any polynomial divisible by  must be divisible by both  and . 

A polynomial is divisible by  if and only if the sum of its coefficients is 0. Add the coefficients for each given polynomial:

:

:

:

The last two polynomials are both divisible by . The other two can be eliminated as correct choices. 

A polynomial is divisible by  if and only if the alternating sum of its coefficients is 0- that is, if every other coefficient is reversed in sign and the sum of the resulting numbers is 0. For each of the two uneliminated polynomials, add the coefficients, reversing the signs of the  and  coefficients:

:

:

The last polynomial is divisible by both  and , and, as a consequence, by .

A polynomial is divisible by  if and only if the sum of its coefficients is 0. Add the coefficients for each given polynomial.

:

:

Since its coefficients add up to 0,  is the only one of the given polynomials divisible by .

Let  be a sixth root of 729. The question is to find a solution of the equation

.

Subtracting 64 from both sides, this equation becomes

729 is a perfect square (of 27) The binomial at left can be factored first as the difference of two squares:

27 is a perfect cube (of 3), so the two binomials can be factored as the sum and difference, respectively, of two cubes:

The equation therefore becomes 

.

By the Zero Product Principle, one of these factors must be equal to 0.

If , then ; if , then . Therefore,  and 3 are sixth roots of 729. However, these are not choices, so we examine the other polynomials for their zeroes.

If , then, setting  in the following quadratic formula:

If , then, setting  in the quadratic formula:

Therefore, the set of sixth roots of 729 is 

Of the choices given,  is the one that appears in this set.