Factor each using the sums of differences of cubes. simplify completely 512x^3 + 125 and 216x^3-64

Question

pleaseee help need the answer quick 

Answer 1

The formula of factoring cube sums is:

x³ + y³ = (x + y)(x² - xy + y²)

The cube root of 512x³ is 8x (8 * 8 * 8 = 512. x * x * x = x3).

The cube root of 125 is 5 (5 * 5* 5 = 125)

So we have our x, we have our y, so let's put them into the equation:

(8x + 5)(64x2 - 8x * 5 + 25)

We have 64 and 25 as part of our equation because the x and y are squared. What is 8² and 5² again? 64 and 25 respectively.

Now we multiply the middle term to finish:

(8x + 5)(64x2 - 40x + 25)

 

For the factoring of cubic differences, we have:

x3 - y3 = (x - y)(x² + xy + y²)

Note the difference of where the subtraction sign is.

the cube root of 216x³ is 6x

The cube root of 64 is 4

So we have:

(6x - 4)(36x² + 6x * 4 + 16)

(6x - 4)(36x² + 24x + 16)

Answer 2

Here are the formulas, as found in a Google search:

a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b)(a2 + ab + b2)

(the numbers are exponents)

 

#1:

512x^3 + 125

 

recognize that 512 and 125 are both cubes, so we can rewrite like so:

(8x)^3 + 5^3

 

Thus--for the formula:  a = 8x, and b = 5

 

Plug into the formula:    <<use the first formula>>

(8x)^3 + 5^3 = (8x + 5) ((8x)^2 - 40x + 25)

simplify:

(8x + 5) (64x^2 -40x + 25)  <<The quadratic term does not appear to be factorable>>

 

 

#2:

216x^3 - 64

rewrite:

(6x)^3 - 8^3

 

So:  a = 6x, and b = 8

Plug these into the second formula