Question

27x^3 − 125 = 0 can somebody show me a working of this equation so that i can compare with mine and see what am not doing right please

Answer 1

do you recognize the difference of 2 cubes?

Answer 2

yes

Answer 3

(3x)^3 - 5^3

Answer 4

okay

Answer 5

i got 3 answers but the computer will not accept the answers

Answer 6

what do you have?

Answer 7

5/3 +-5sqt 675/18

Answer 8

why do you have sqrt when doing cubes?

Answer 9

a^3 -b^3 = (a - b)(a^2 + ab + b^2)

Answer 10

It equals 0, though. You only need to isolate x, you shouldn't need to separate out the difference

Answer 11

(3x-5){ (3x)^2 + 2(3x) + 5^2 }

Answer 12

27x^3 - 125 = 0
You can do algebra to start the process of isolation. Simply add 125 to both sides:
27x^3 = 125
And then divide both sides by 27:
x^3 = 125/27
Am I correct in assuming that you know how to take cube roots?

Answer 13

yes please

Answer 14

even thought i know the quadratic better

Answer 15

Alright, so, have you ever used the property of square roots that looked like:
\[\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\]? The same logic applies to cube roots.

Answer 16

yes just did last week

Answer 17

\[\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}\] We can use this property in the equation we need to solve, \[x^3 = \frac{125}{27}\]

Answer 18

To undo x^2, you take the square root of both sides. Likewise, to undo x^3, you take the cube root of both sides (the square root sign with the little 3 next to it).

Answer 19

\[\sqrt{x^2}=x \rightarrow \sqrt[3]{x^3}=x\] Square roots find the number that, when multiplied by itself (x*x, or x^2), equals the number you are taking the square root of. Cube roots find the number that, when multiplied by itself 3 times (x*x*x, or x^3), equals the number you are taking the cube root of.

Answer 20

For example, we know that 2^3 = 2*2*2 = 8, so that means the cube root of 8 is 2.

Answer 21

It's handy to memorize the cubes of small integers like you probably memorized the squares of small integers.

1^3 = 1
2^3 = 8
3^3 = 27
4^3 = 64
5^3 = 125
6^3 = 216
7^3 = 343
8^3 = 512
9^3 = 729
10^3 = 1000

As you can see, they get pretty big very quickly... but we can use these in our problem, particularly the fact that 3^3=27 and 5^3=125.

Answer 22

@Tomhue's method will only give you one value for x. You will not get the complete solution that includes the values with i .

Answer 23

Let's take the cube root of both sides of what we got to:
\[x^3 = \frac{125}{27} \rightarrow \sqrt[3]{x^3}=\sqrt[3]{\frac{125}{27}}=\frac{\sqrt[3]{125}}{\sqrt[3]{27}}\]

Answer 24

You want roots of unity? I can do roots of unity :p

Answer 25

So, let's look at our algebra again. Here are all the steps from start to finish (with cbrt(x) denoting the cube root of x):
27x^3 - 125 = 0
27x^3 = 125
x^3 = 125/27
cbrt(x^3) = cbrt(125/27)
x = cbrt(125)/cbrt(27)

Remember how we saw that 5^3=125 and 3^3=27? That means cbrt(125)=5 and cbrt(27)=3.
x = 5/3
This is the first solution.If you're in Algebra 1, 2, or Geometry, I think this is probably the only solution you need. Have you learned about the complex plane yet?

Answer 26

am actually taking college Algebra 1314

Answer 27

i got 5/3 in my first equation too

Answer 28

yes that's correct

Answer 29

Alright, then you might need the other two answers. Think about roots as points on a circle in the complex plane.

Remember that square roots can have two answers, a positive and a negative value, so you have to write +/- sqrt(x) sometimes. The two answers are exactly opposite each other on the circle -- they are equidistant, angle-wise. (They're 360/2 = 180 degrees apart, or 2pi/2 = pi radians apart.)

Cube roots are similar, but they have three total answers rather than two total answers. The three points on the circle will be 360/3 = 120 degrees apart, or 2pi/3 radians apart. Now, calculating these involves some trig calculations that frankly I do not feel like going through at 12:24 a.m.

The positive root we got is called the "principal root" and is usually will suffice as an answer.

However, if you do need the other two roots, you can use some trig (actually just right triangle calculations) to figure out where the other two are on the complex plane.