factorise 8z^౩ -(x-y)^3-(y+z)3-(z-x)౩

Question

please solve

Answer 1

We have this: 8z³ -(x-y)³-(y+z)³-(z-x)³

 

First let's recall the sum/difference of cubes formulas:

 

a³ - b³ = (a - b)(a² + ab + b²)

a³ + b³ = (a + b)(a² - ab + b²)

 

You should see that the first two cubes and the last two cubes in your expression have (2z - x + y) as a factor as follows:

 

8z³ - (x-y)³

= [2z - (x - y)][4z² + 2z(x - y) + (x - y)²]

= (2z - x + y)(x² - 2xy + 2xz + y² - 2yz + 4z²)

 

(y+z)³ + (z - x)³

= [(y + z) + (z - x)][(y + z)² - (y + z)(z - x) + (z - x)²]

= (2z - x + y)(x² + xy - xz + y² + yz + z²)

 

Now the original expression equals the difference of these two factorizations, so we can combine with the distributive property.

 

(2z - x + y)(x² - 2xy + 2xz + y² - 2yz + 4z²) - (2z - x + y)(x² + xy - xz + y² + yz + z²)

=(2z - x + y)(-3xy + 3xz - 3yz + 3z²)

 

The second factor is handled by grouping.

 

-3xy + 3xz - 3yz + 3z²

= -3[(xy - xz) + (yz - z²)]

= -3[x(y - z) + z(y - z)]

= -3(x + z)(y - z)

 

Finally, we can distribute the negative sign in the -3 into the original factor we found and get the final answer.

 

3(x - y - 2z)(x + z)(y - z)