Question

factor completely:
c^6-27

Answer 1

\[c^6-27 \implies (c^2)^3-(3)^3\]Now can u use this identity here\[a^3-b^3=(a-b)(a^2+ab+b^2)\]

Answer 2

a=c^2
b=3

Answer 3

(c^2-3)(c^4+3c^2+9)

Answer 4

yup ! that's ur answer :)

Answer 5

what is that formula called?

Answer 6

Nothing it's just an algebraic identity :)

Answer 7

There are many algebraic identities through which ur factoring will become easy :)

Answer 8

Do u want those identities ?

Answer 9

yes

Answer 10

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