Factoring Differences of Cubes: x^12 - y^3 z^12

[ReginaR]

I am limping along in my homework and this equation has me stumped.
x^12 - y^3 z^12

This is my initial mind frame to solve it

((x^4)^3 - y^3 (z^4)^3)
Where A = x^4 and B = y^3 z^4

I know that this is probably incorrect, but can someone help me please? I do not want the answer only , but the steps so that I can understand the process.

Thanks in advance

 

[stapel]

I am limping along in my homework and this equation has me stumped.
x^12 - y^3 z^12

This is my initial mind frame to solve it

((x^4)^3 - y^3 (z^4)^3)
Where A = x^4 and B = y^3 z^4

I'm not sure how you're using "A" and "B" here...? Instead, try finding the cube root of each of the first and second terms.

. . . . .\(\displaystyle x^{12}\)

What would you cube to get this? (Yes, you're right: x⁴.)

. . . . .\(\displaystyle y^3\, z^{12}\)

What would you cube to get this? (No, not y³ z⁴.)

Then plug these into the difference-of-cubes formula.

 

[Steven G]

I am limping along in my homework and this equation has me stumped.
x^12 - y^3 z^12

This is my initial mind frame to solve it

((x^4)^3 - y^3 (z^4)^3)
Where A = x^4 and B = y^3 z^4

I know that this is probably incorrect, but can someone help me please? I do not want the answer only , but the steps so that I can understand the process.

Thanks in advance

y^3 (z^4)^3 =(yz^4)^3. So now you have the difference of cubes. Continue.

 

[hoosie]

Factorising Difference of Two Cubes

Factorise the difference of two cubes:
x^12 - y^3 z^12


a³ - b³ = (a-b)(a²+ab+b²)
where
a = x^4 since x^4 * x^4 * x^4 = x^12
b = yz^4 since yz^4 * yz^4 * yz^4 = y^3 z^12


It follows:


x^12 - y^3 z^12
= (x^4 - yz^4){(x^4)^2 + x^4*yz^4 + (yz^4)^2}
= (x^4 - yz^4)(x^8 + x^4yz^4 + y^2z^8)


Check: let x = 2, y = 1, z = 3
LHS = 2^12 - 1^3*3^12
= 4096 - 1*531 441
= -527 345
RHS = (2^4 - 1*3^4)(2^8 + 2^4*1*3^4 + 1^2*3^8)
= (16 - 81)(256 + 1296 + 6561)
= -65 * 8113
= -527 345
= LHS
Therefore answer is correct.

 

[Steven G]

Factorise the difference of two cubes:
x^12 - y^3 z^12


a³ - b³ = (a-b)(a²+ab+b²)
where
a = x^4 since x^4 * x^4 * x^4 = x^12
b = yz^4 since yz^4 * yz^4 * yz^4 = y^3 z^12


It follows:


x^12 - y^3 z^12
= (x^4 - yz^4){(x^4)^2 + x^4*yz^4 + (yz^4)^2}
= (x^4 - yz^4)(x^8 + x^4yz^4 + y^2z^8)


Check: let x = 2, y = 1, z = 3
LHS = 2^12 - 1^3*3^12
= 4096 - 1*531 441
= -527 345
RHS = (2^4 - 1*3^4)(2^8 + 2^4*1*3^4 + 1^2*3^8)
= (16 - 81)(256 + 1296 + 6561)
= -65 * 8113
= -527 345
= LHS
Therefore answer is correct.

Nice work. I only object with your last line saying that therefore the answer is correct. Just because your check works does not mean that your answer is correct. It just means that you can't say that you're wrong.