How To Factor Perfect Cube

Factoring the Sum and Difference of Two Cubes

In algebra class, the teacher would always discuss the topic ofsum of two cubes and divergence of two cubes next. The reason is that they are similar in construction. The cardinal is to "memorize" or remember the patterns involved in the formulas.

Case ane: The polynomial in the grade ${a^iii} + {b^3}$ is called the sum of ii cubes considering two cubic terms are existence added together.

Case ii: The polynomial in the form ${a^3} - {b^3}$ is called the difference of ii cubes considering 2 cubic terms are being subtracted.

Then here are the formulas that summarize how to factor the sum and difference of two cubes. Written report them carefully.

Case 1: Sum of Two Cubes

Observations:

For the "sum" case, the binomial factor on the right side of the equation has a middle sign that is positive.

In addition to the "sum" case, the eye sign of the trinomial factor volition always be contrary the middle sign of the given problem. Therefore, it is negative.

Case 2: Deviation of Two Cubes

Observations:

For the "deviation" instance, the binomial factor on the right side of the equation has a middle sign that is negative.

In addition to the "divergence" case, the heart sign of the trinomial gene will ever be opposite the middle sign of the given problem. Therefore, it is positive.

Examples of How to Factor Sum and Departure of Ii Cubes

Let's go over some examples and come across how the rules are applied.

Case 1: Factor ${10^3} + 27$ .

Currently, the problem is non written in the form that we want. Each term must exist written as a cube, that is, an expression raised to a power of $iii$ . The term with variable $x$ is okay but the $27$ should be taken care of. Obviously, we know that $27 = \left( three \right)\left( three \right)\left( 3 \correct) = {3^3}$ .

Rewrite the original problem as sum of two cubes, and and then simplify. Since this is the "sum" case, the binomial factor and trinomial factor volition have positive and negative center signs, respectively.

Instance 2: Factor ${y^iii} - 8$ .

This is a instance of difference of two cubes since the number $8$ can be written as a cube of a number, where $viii = \left( 2 \right)\left( ii \correct)\left( 2 \correct) = {ii^3}$ .

Apply the rule for divergence of two cubes, and simplify. Since this is the "departure" example, the binomial cistron and trinomial factor volition take negative and positive middle signs, respectively.

Example three: Factor $27{x^3} + 64{y^iii}$ .

The first step as always is to express each term as cubes. We know that $27 = {three^3}$ and $64 = {4^iii}$ . Rewrite the trouble as sum of two cubic terms and utilize the dominion, so we become

Instance 4: Factor $125{ten^three} - 27$ .

Since $125 = \left( 5 \right)\left( v \right)\left( 5 \correct) = {5^3}$ and $27 = \left( 3 \right)\left( three \right)\left( 3 \correct) = {three^three}$ , this is conspicuously a problem on difference of 2 cubes.

Here's the solution.

Example 5: Factor $1 - 216{x^iii}{y^three}$ .

At first, this problem may look "difficult". However, if you stick to what we know already nearly sum and difference of two cubes nosotros should be able to recognize that this problem is rather easy.

The proficient thing is that the variables are cubes then they are fine. At present for the number, it is piece of cake to see that that $1 = \left( 1 \correct)\left( 1 \right)\left( 1 \right) = {i^3}$ while $216 = \left( 6 \right)\left( half-dozen \correct)\left( half-dozen \right) = {6^iii}$ . This is actually a case of departure of two cubes.

Example vi: Factor $8{x^6}{y^{12}} + 27$ .

This problem is a bit unlike. The coefficients are definitely cubes because $8 = {2^3}$ and $27 = {3^three}$ . Now, how practise we limited the term with variables as a cube? Well, simply gene out $3$ from the existing exponents of " $x$ " and " $y$ ". Use the law of exponent known as Power to a Power Dominion to justify this step. We take out $three$ because to exist a cube implies that any expression must have an outer exponent of $iii$ .

8x^6y^12+27=(2x^2y^4+3)(4x^4y^8-6x^2y^4+9)

Example 7: Gene $3xy - 24{ten^iv}y$ .

Sometimes the problem may not appear to be factorable by either sum or difference of two cubes. If you run across something similar this, try to take out mutual factors. For the numbers, the greatest common factor is $3$ and for the variables, the greatest common factor is " $xy$ ". Therefore the overall mutual cistron would exist their product which is $\left( three \right)\left( {xy} \correct) = 3xy$ .

After factoring it out, y'all'll see that nosotros take an like shooting fish in a barrel problem on the difference of two cubes.