Expression

[gorka09]

I've done something similar to this, but I'm not sure if I can do it the same way. The expression is:

a^6 - b^6
__________
a-b


Normally, I would subtract 6 (a exponent on top) from 1 ( a exponent on bottom), and then do the same for b, but I don't think that is what you do. Any help is appreciated!

 

[skeeter]

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

now ... let x = a^2 and y = b^2

 

[gorka09]

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

now ... let x = a^2 and y = b^2

So do you distribute the x-y, and the solution to that would equal x?

 

[skeeter]

no ... the pattern shown was given to help you factor a^6 - b^6

(a^6 - b^6)/(a - b) = (a^2 - b^2)(a^4 + a^2*b^2 + b^4)/(a - b)

now ... what can you do with a^2 - b^2 on the right side of the equation?

 

[stapel]

So do you distribute the x-y...?

No; multiplying the (x - y) factor back through would take you right back where you'd started. Use the factorization that the tutor gave you, and cancel.

...and the solution to that would equal x?

The solution to what would be "x"? Unless much of the exercise has been omitted, you're not "solving" anything; you're simplifying an expression.

By replacing a² with x and b² with y, the tutor related the given expression, a⁶ - b⁶ = (a²)³ - (b²)³, to the simpler difference-of-cubes formula you've memorized. Just plug into that.

Eliz.

 

[gorka09]

What do you mean "plug in"?

Would the answer be (a-b)(a^2+ab+a^2)?

 

[Gene]

"Plug in" means "replace by."

x^3 - y^3 = (x - y)(x^2 + xy + y^2)

now ... let x = a^2 and y = b^2

Replace x by a^2 and y by b^2 to get
(a^2-b^2)((a^2)^2 - a^2*b^2 +(b^2)^2) =
(a^2-b^2)(a^4 - a^2*b^2 +b^4)
as Skeeter said.

Leave
(a^4 - a^2*b^2 +b^4)
alone. It is part of the answer just as it is.

Look at
(a^2 - b^2)/(a-b)
Simplify that then put that answer in front of the
(a^4 - a^2*b^2 +b^4) for the final
answer.