How do you factor #z^6-64p^6#?

Answer 1

Eva Abramson

#z^6-64p^6=(z-2p)(z^2+2pz+4p^2)(z+2p)(z^2-2pz+4p^2)#

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

The sum of cubes identity can be written:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Hence:

#z^6-64p^6#

#=(z^3)^2-(8p^3)^2#

#=(z^3-8p^3)(z^3+8p^3)#

#=(z^3-(2p)^3)(z^3+(2p)^3)#

#=(z-2p)(z^2+2pz+4p^2)(z+2p)(z^2-2pz+4p^2)#

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Answer 2

Savannah Anderson

To factor ( z^6 - 64p^6 ), you can use the difference of squares formula, which states that ( a^2 - b^2 = (a + b)(a - b) ). Applying this formula, you get ( (z^3)^2 - (4p^3)^2 ). Then, you can factor it into ( (z^3 + 4p^3)(z^3 - 4p^3) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.