Question

Need help to factorise 2m^12-1/8?

Answer 1

\[2m^{12}-\tfrac18\]

its a difference of cubes

Answer 2

how is this n cube?

Answer 3

do you know the cube root of eight?

Answer 4

i think its 2?

Answer 5

yes


so the cube root of 1/8 is ?

Answer 6

1/2?

Answer 7

that's right

Answer 8

now can you take the cube root of \( m^{12}\) ?

remember that \[\sqrt[3]x=x^{1/3}\]and \[(y^a)^b=y^{ab}\]

Answer 9

12root of m, i dont know how to make the root sign. sorry

Answer 10

[alt][v] √

Answer 11

for cube root ^3√

Answer 12

so what is

^3√(m^12) = (m^12)^(1/3)

Answer 13

give me a moment to think about this, its long time since i did any cube functions

Answer 14

A bit simpler?
\[2m^{12}-\frac{ 1 }{ 8 }=2(m^{12}-\frac{ 1 }{ 16 })=2(m^6+\frac{ 1 }{ 4 })(m^6-\frac{ 1 }{ 4 })=\]\[=2(m^6+\frac{ 1 }{ 4 })(m^3+\frac{ 1 }{ 2 })(m^3-\frac{ 1 }{ 2 })\]

Answer 15

2m^12-1/8 = 2 ( m^12 - 1/16)

then you get:
2 ( m^12 - 1/16) = 2 (m^6 1/4) (m^6 - 1/4)
= 2 (m^6 1/4) (m^3 1/2) (m^3 - 1/2)

this is what i originally got but was unsure because i did this stuff along time ago, so i just wanted to check with you guys cause my confidence in maths is not there yet

Answer 16

That is correct

Answer 17

CarlosGP : how do you get it to type your factors like that, i have no clue

Answer 18

use the button:\[\sum_{}^{}Equation\] (bottom left of the text editor)

Answer 19

```
\[\frac{a+b}{d}\]
```

Answer 20

thanks you guys

Answer 21

u r welcome