Question

(1/8)x^3-(1/27)y^3

Answer 1

hint: this is difference of two cubes
\[\huge (\frac x2)^3 - (\frac y3)^3\]
does that help?

or do you need more help?

Answer 2

im confused, how does that apply here? sorry im bad at alg 2..

Answer 3

did you know that the formula for difference of two cubes is \[\huge a^3 - b^3 \implies (a-b)(a^2 + ab + b^2)\]

Answer 4

so...what do you think would be the factored form for your problem?

Answer 5

(1/8x^3-1/27y^3)(1/8x^6+1/208xy^4+1/27y^6?

Answer 6

no no... remember i said you turn 1/8 x^3 - 1/27 y^3 into \[\huge (\frac x2)^3 - (\frac y3)^3\]

Answer 7

so your a here is x/2 and b is y/3

got it?

Answer 8

then where do the 1/8 and 1/27 go?

Answer 9

1/8 = (1/2)^3

1/27 = (1/3)^3

Answer 10

i factored it out

Answer 11

\[\huge \frac{x^3}{8} \implies \frac{x^3}{2^3} \implies (\frac x2)^3\]
\[\huge \frac{y^3}{27} \implies \frac{y^3}{3^3} \implies (\frac y3)^3\]
see?

Answer 12

OH thats where the denominators came from..

Answer 13

yep

Answer 14

so now... can you factor (x/2)^3 - (y/3)^3?

Answer 15

1/216 (3x-2y)(9x^2+6xy+4y^2) ?

Answer 16

wait what? what happened?

Answer 17

i think i need to show you an example....

Answer 18

please do

Answer 19

\[125x^3 - 216y^3\]
first... i turn this into a^3 - b^3 form so...
\[\implies (5x)^3 - (6y)^3\]
now... i put it in (a-b)(a^2 + ab + b2)..my a here is 5x and b is 6y so..
\[\implies (5x - 6y)(25x^2 + 60y + 36y^2)\]

Answer 20

(x/2)^3 - (y/3)^3, (x/2-y/3)(x^2/4+xy/6+y/9)?

Answer 21

yes!!! just get rid of that (x/2)^3 - (y/3)^3 at the start

Answer 22

i just had to paste it there to remember what i started with haha i meant to delete it after, thank you so so much youre very patient hahah

Answer 23

welcome