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.

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

where the  terms are the four zeroes.

A polynomial with rational coefficients has its imaginary zeroes in conjugate pairs. Since  is such a polynomial, then, since  is a zero of multiplicity 2, so is its complex conjugate . We can set  and , and 

We can rewrite this as

or

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

Therefore, 

Multiplying:

This equation cannot be factored.  Since this is a parabola, we can use the quadratic equation to find the roots.

Substitute the coefficients of .

The answers are:  

One can find the points of intersection of these two functions by setting them equal to one another, essentially substituting  in one equation for the -side of the other equation. This will tell us when the  (output) will be the same in each equation for a given  (input).

By simplifying this equation and setting it equal to zero, we can find the two -values that produce the same  values in the system of the two equations.

Subtract from both sides of the equation, and add to both sides.

Factoring this last equation makes it easier to find the -values that will result in zero on the left side of the equation. Set the two parenthetical phrases equal to zero to find two separate -values that satisfy the equation. These  values will be the  values of the points of intersection between the two equations.

We know our factors multiply to , and the six times one factor plus the other is equal to

and , so and are our factors.

Substituting these two values into either of the two original equations results in the -values of the points of intersection.

 are the points of intersection.

To solve this system of equations, we must first eliminate one of the variables. We will begin by eliminating the  variables by finding the least common multiple of the  variable's coefficients.  The least common multiple of 3 and 2 is 6, so we will multiply each equation in the system by the corresponding number, like

.

By using the distributive property, we will end up with

Now, add down each column so that you have

Then you solve for  and determine that .

But you're not done yet!  To find , you have to plug your answer for  back into one of the original equations:

Solve, and you will find that .  

Rewrite the two equations by setting  and  and substituting:

The result is a two-by-two linear system in terms of  and :

This can be solved, among other ways, using Gaussian elimination on an augmented matrix:

Perform row operations until the left two columns show identity matrix . One possible sequence:

 and .

In the former equation, substitute  back for :

.

Taking the reciprocal of both sides, we get

.

Rewrite the two equations by setting  and  and substituting:

The result is a two-by-two linear system in terms of  and :

This can be solved, among other ways, using Gaussian elimination on an augmented matrix:

Perform row operations until the left two columns show identity matrix . One possible sequence:

 and .

Substituting back in the second equation and solving for  by cubing both sides:

Rewrite the two equations by setting  and  and substituting:

In terms of  and , this is a two-by-two linear system: 

Rewrite this as an augmented matrix as follows:

Perform the following row operations to make the left two columns the identity matrix :

 and .

Replacing  with , then squaring both sides:

To solve for , we will need to eliminate the  variable.  To do so,  we can choose either the substitution or the elimination method.  

Let's choose the substitution method.  Rewrite the second equation so that  is isolated.

Subtract  on both sides.

Divide by five on both sides.

Substitute this back to the first equation.

Distribute the terms.

Multiply both sides by 5 to eliminate the fractions.

Add 8 on both sides and combine like-terms on the left.

Divide by 18 on both sides.

The answer is:  

In order to solve for the x-variable, we will need to use the elimination method to eliminate the y-variable.

Multiply the second equation by two.

Add both equations.

Divide by five on both sides.

The answer is: