Question

Factor this expression completely, then place the factors in the proper location on the grid. 1/8x^3-1/27y^3

Answer 1

\[(a-b)^3=(a-b)(a^2+ab+b^2)\]
Find your a and b and use the formula.

Answer 2

can you answer ?

Answer 3

i give medal

Answer 4

can you tell what a and b is? from your equation?

Answer 5

\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 6

look on screen

Answer 7

i understand the problem, but i'm helping you do it, not doing it for you friend.

Answer 8

im stuck + this was last question of the day

Answer 9

solve please

Answer 10

its not hard, i can help you... can you tell me what a^3 is?

Answer 11

a squared 3

Answer 12

from the formula above, you need to find a and b using cube roots

Answer 13

i hate fractions doing in factor

Answer 14

\[\frac{ 1 }{ 8 }x^3-\frac{ 1 }{ 27 }y^3\]

this is the formula for factoring:

\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

You need to find the a and the b. what times itself 3 times gives 8 and what times itself gives 27?

Answer 15

then pretend they're not fractions, just use the denominator for now pretend its 8x^3-27y^3

Answer 16

what is the cube root of 8 and what is the cube root of 27?

Answer 17

The cube root of 8 is 2

Answer 18

good

Answer 19

for 27 is 3

Answer 20

good. so use the formula with a=2x and b=3y

\[a^3-b^3=(a-b)(a^2-ab-b^2)\]

\[8x^3-27y^3=(2x-3y)((2x)^2+(2x)(3y)+(3y)^2)\]

The only difference with your problem is the numbers will be denominators with 1 as the numerator.

Answer 21

i have to put 1/6x in grid with they answer

Answer 22

yes the answer will have a 1/6 in it somewhere

Answer 23

actually it should be 1/6xy

Answer 24

1/216 (3x-2y)(9x^2+6xy+4y^2) that's what comes up to me

Answer 25

\[8x^3−27y^3=(2x−3y)((2x)^2+(2x)(3y)+(3y)^2)\]

\[(2x−3y)((2x)^2+(2x)(3y)+(3y)^2)\] is the part that needs calculating. The first two terms are done, now what is the first term of the second part of the parenthesis (2x)^2?

Answer 26

4x^2?

Answer 27

yes... so what's the rest?

Answer 28

(4x^2 + 6xy + 9y^2) is the last part...

therefore

\[8x^3-27y^2=(2x-3y)(4x^2+6xy+9x^2)\]

the only difference is now all the numbers are under the fraction with a 1 on top.

Answer 29

8^3-6xy-27Y^3=9y^2

Answer 30

?

Answer 31

\[\frac{ 1 }{ 8}x^3-\frac{ 1 }{ 27}y^3=(\frac{ 1 }{ 2}x-\frac{ 1 }{ 3}y)(\frac{ 1 }{ 4}x^2+\frac{ 1 }{ 6}xy+\frac{ 1 }{ 9}y^2)\]

Answer 32

thats the answer?

Answer 33

yeah

Answer 34

their is no 1/8 in box look

Answer 35

its the problem and the answer together

Answer 36

i need put answer on grid can you type your answer so it will fit the gird look on screen

Answer 37

ok the right of the equal sign is the answer, the left of the equal sign is the orginal problem

Answer 38

ok got it :)

Answer 39

are you teacher? in past or right now :) your good explaining

Answer 40

yes i am just here to help people

Answer 41

thanks

Answer 42

its just a formula you plug the numbers in claylordmath, you're making it way harder than it is. i know it looks scary but its not really that hard.