Question

can someone please help me with factor 27x^3-8y^3 i no 9 and 3 is 27 and -4 and 2 is -8

Answer 1

ok - I assume you know how to factorise this expression:\[a^3-b^3\]

Answer 2

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

Answer 3

not quite, it should be:\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 4

so now you just need to express your equation as the difference of two cubes

Answer 5

your first term is \(27x^3\) - how can this be expressed as (something)^3 ?

Answer 6

3^3 and 2^3

Answer 7

?
\[27x^3=(???)^3\]what should the ??? be replaced by?

Answer 8

3

Answer 9

3x

Answer 10

correct, so the first term can be expressed as \((3x)^3\)
do the same for the second term - what do you get?

Answer 11

(2y)^3

Answer 12

perfect, so your original expression can be written as:\[27x^3-8y^3=(3x)^3-(2y)^3\]now it is in the form where you can make use of the factorisation:\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 13

3^3-2y^3= (3-2) (3^2+3*4+4^2)

Answer 14

???

Answer 15

what did you use to represent 'a' and 'b' ?

Answer 16

3 and 2

Answer 17

remember we ended up with:\[27x^3-8y^3=(3x)^3-(2y)^3\]and we know:\[a^3-b^3=(a)^3-(b)^3=(a-b)(a^2+ab+b^2)\]so look carefully at what 'a' and 'b' should be.

Answer 18

a is 3 and 2 is b

Answer 19

what expression is inside the braces for the first term in: \((3x)^3-(2y)^3\) ?

Answer 20

3x and 2y

Answer 21

well - first term is 3x. so we know 'a' must be '3x'
similarly, 2nd term is '2y'. so we know 'b' must be '2y'

Answer 22

do you understand?