Question

81w^6-3p^6
factor
medals given

Answer 1

lol not again...this is based on the same formula..why not try it urself ?

Answer 2

I have tried. I just can't get the right answer. All the ones I have posted I have done several times on my own.

Answer 3

take the 3 out

Answer 4

look for common factors first, you finish it

Answer 5

3(27w^6-p^6)

Answer 6

this is why you should be familiar with the squares and cubes of the numbers 1-12

Answer 7

so its easy to find patterns.

Answer 8

or

\[(9w^3-\sqrt{3}p^3)(9w^3+\sqrt{3}p^3)\]

Answer 9

3(3w^2-p^2)(9w^4+3w^2*p^2+p^4)
thats it

Answer 10

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

Answer 11

even then that first difference of sqaures should be factored further

Answer 12

you also have
a different of cubes formula
and a sum of cubes formula

Answer 13

3w^2 -p^2 , shouldnt leave as this, everything should be linear or irreduciable quadratic.

Answer 14

whats the problem with my answer then?
\[a ^{3}-b ^{3}=(a ^{2}+ab+b ^{2}) (a-b)\]

Answer 15

thanks guys. all these formulas! I can't keep up.