Question

HELP!!

Can anyone solve this??
1) x^8 - y^8
2) 9x^4 - 16y^8
3) 8x^6 - 27
4) 4x^2/9 - y^4
* Difference of Squares & Sum or Difference of Cubes

Answer 1

1.) x^8 - y^8 = (x4 + y4) • (x4 - y4) - squares
2.) \[3^2x^4 - 2^4y^8\] - squares
3.) 8x^6-27 - cubes
4.) I'm not sure on this one. Is 2/9 a fraction or is it dividing the term 4x^2 by 9? If I had to guess, it would be squares.

Answer 2

Thank you for your answering and... yes it is divided by 9 in no. 4

Answer 3

is it really minus in squares on no. 1??? Im confused

Answer 4

1) can be factored more
the expression (x^4-x^4) is also a difference of squares
2) seems a bit off but yes use difference of squares
3) agreed used difference of cubes
4) use difference of squares

Answer 5

Can you do solve no. 4??

Answer 6

\[u^2-v^2=(u-v)(u+v) \text{ is how to factored difference of squares } \\ u^3+v^3=(u+v)(u^2-uv+v^2) \text{ is how to factored sum of cubes } \\ u^3-v^3=(u-v)(u^2+uv+v^2) \text{ is how to factored difference of cubes }\]

Answer 7

\[\frac{a^2}{b^2}x^2-v^2 \\ \text{ where } v=u^2 \\ \text{ first all notice this can be written as } \\ (\frac{a}{b}x)^2-v^2 \\ \text{ this is a difference of squares }\]

Answer 8

example:
\[\frac{16}{25}x^2-u^4 \\ \frac{4^2}{5^2}x^2-(u^2)^2 \\ (\frac{4}{5}x)^2-(u^2)^2 \\ (\frac{4}{5}x-u^2)(\frac{4}{5}x+u^2)\]

Answer 9

\[\frac{4}{9}x^2-y^4 \\ \frac{2^2}{3^2}x^2-(y^2)^2 \\ (\frac{2}{3}x)^2-(y^2)^2 \]
can you try the factoring part?

Answer 10

it is just pluggin into a formula really

Answer 11

Okay, i'll try that...

Answer 12

Thank you so muuuch!!

Answer 13

@freckles