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

Question

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

take the 3 out

anonymous · Answer

look for common factors first, you finish it

myininaya · Answer

3(27w^6-p^6)

anonymous · Answer

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

anonymous · Answer

so its easy to find patterns.

myininaya · Answer

or

$$(9w^3-\sqrt{3}p^3)(9w^3+\sqrt{3}p^3)$$

anonymous · Answer

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

myininaya · Answer

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

anonymous · Answer

even then that first difference of sqaures should be factored further

myininaya · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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