Answer:
<u>Equation:</u>
Step-by-step explanation:
<u>Step 1:</u>
- Pull out like factors:

<u>Trying to factor as a Difference of Cubes:</u>
- Factoring:

- Theory : A difference of two perfect cubes, a^3 - b^3 can be factored into
- (a-b) • (a^2 +ab +b^2)
- Proof : (a-b)•(a^2+ab+b^2) =
- a^3+a^2b+ab^2-ba^2-b^2a-b^3 =
- a^3+(a^2b-ba^2)+(ab^2-b^2a)-b^3 =
- a^3+0+0-b^3 =
- a^3-b^3
- Check : g^1 is not a cube !!
- Ruling : Binomial cannot be factored as the difference of two perfect cubes
<u>Equation at end of step 1:</u>
- <u />

<u>Step 2:</u>
- A product of several terms equals zero.
- When a product of two or more terms equals zero, then at least one of the terms must be zero.
- We shall now solve each term = 0 separately
- In other words, we are going to solve as many equations as there are terms in the product
- Any solution of term = 0 solves product = 0 as well.
<u>Solving a Single Variable Equation:</u>
- Solve

- In this type of equations, having more than one variable (unknown), you have to specify for which variable you want the equation solved.
- We shall not handle this type of equations at this time.
<u>Solution:</u>
Answer:
The answer is 180.
Just multiply thirty and six.
Hope that helps!
The correct answer is 58 yes
Answer:
Choice A. √x
Step-by-step explanation:
In exponents,
![\\ \tt \: {a}^{ \frac{1}{b} } = \sqrt[b]{a}](https://tex.z-dn.net/?f=%20%20%5C%5C%20%20%5Ctt%20%5C%3A%20%20%7Ba%7D%5E%7B%20%5Cfrac%7B1%7D%7Bb%7D%20%7D%20%20%3D%20%20%5Csqrt%5Bb%5D%7Ba%7D%20)
Hence,
![\\ \tt \: {x}^{ \frac{1}{2} } = \sqrt[2]{x} \: \: \: or \: \: \: \sqrt{x}](https://tex.z-dn.net/?f=%20%5C%5C%20%20%5Ctt%20%5C%3A%20%20%20%7Bx%7D%5E%7B%20%5Cfrac%7B1%7D%7B2%7D%20%7D%20%20%3D%20%20%5Csqrt%5B2%5D%7Bx%7D%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20or%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5Csqrt%7Bx%7D%20)
Answer:
.um g o o g l e look it u p I do that.