how do you factor: z^3 + 64?

To factor a sum or difference of cubes, do the following steps.

1) Take the cube root of each term; place in the first set of brackets.

In the second set of brackets, you will have three terms:

2) "Square the first" - square the first term in the first set of brackets.
3) "Negative product" - multiply the two terms together, and then take their negative (multiply it by -1).
4) "Square the last" - square the last term in the first set of brackets.

Let's do this one by one, shall we?

z^3 + 64

Take the cube root of each. The cube root of 64 is 4.

(z + 4)(? + ? + ?)

"Square the first"

(z + 4) (z^2 + ? + ?)

"Negative product". The product of z and 4 is 4z, negated is -4z.

(z + 4)(z^2 - 4z + ?)

"Square the last"

(z + 4)(z^2 - 4z + 16)

initially, the version of squares (ex: a² - b² = (a + b)(a - b)) is something you may desire to comprehend for the examination, memorizing helps nonetheless. you in basic terms gotta comprehend that a² - b² is in basic terms a² + ab - ab - b² once you escalate the (a + b)(a - b). And to respond to your question, to element thoroughly is "to element it as much as feasible so as that what you have in front of you may desire to in basic terms be divisible by utilising a million and itself, and which you will now no longer persist with any form of factoring formulation feasible" ..for a loss of extra effective words... xy² - xz² you are able to attempt "dividing by utilising x", so which you will fsho element x x(y² - z²) Now you got here upon the version of squares. persist with the formulation x(y + z)(y - z) Now to procure that long expression, attempt utilising any formulation to "sparkling it up" extra....you cant! so consequently that is what we call, thoroughly factored. it quite is the sparkling-ist it may get. comparable for b) 8x² - seventy two 8 seems "consumer-friendly" 8(x² - 9) yet observe how 9 is largely 3² good? Rewriting it supplies us: 8(x² - 3²) 8(x + 3)(x - 3) ---Edit: are you able to added element the stuff interior the parenthesis? Nope, so which you're thoroughly factored. comparable factor for c) 64z - z^3 a million z on the sixty 4 and 3 z's on the 2d term, now pull out a z from the two to get... z(sixty 4 - z²) Now keep in mind sixty 4 = 8² z(8² - z²) z(8 + z)(8 - z) Now are you able to do any factoring for the z? no, so which you're all solid to flow right here, that is thoroughly factored and you are able to't quite do something at this element cuz you're finished :D wish this helps and robust success :D

a^3 + b^3 ≡ (a + b)(a^2 - ab + b^2)

z^3 + 64
= z^3 + 4^3
= (z + 4)(z^2 - 4z + 16)

z^3 + 64 is the sum of two cubes. (64 = 4^3)
Generically, a^3 + b^3 can be factored as:

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

In this case:

z^3 + 4^3 = (z+4) ( z^2 - 4z + 4^2) = (z+4) ( z^2 - 4z + 16)

64=4^3..
z^3 + 4^3 (its of the form a^3 + b^3=(a+b)(a^2 - ab + b^2)

=> z^3+4^3=(z+4)(z^2-4z+16)
again factorise the 2nd term

a^3+b^3 = (a+b)(a^2 – ab + b^2)

z^3 + 64=z^3 + (4)^3
=(z+4)(z^2 - 4z + 16)

This is the sum of two cubes A^3 + B^3
The standard factorization is (A + B)(A^2 - AB + B^2)
Or in your case:=

(Z + 4)(Z^2 - 4Z + 16)

This is a sum of squares.

(cube root + cube root)(square - product + square)

(z+4)(z^2-4z+16)

do you remember (x^3 + a^3) =
(x + a)(x^2 -ax +a^2)

x^3+4^3= (x +4)(x^2 -4x + 16)

z³ + 64
= z³ + 4³
= (z + 4)(z² - 4z + 4²)
= (z + 4)(z² - 4z + 16)