Expanding (y^3 + 8)(y^3 - 1) help?

Question

What are the steps to expanding (y^3 + 8)(y^3 - 1) ? I know the answer is (y +2)(y^2 - 2y +4)(y - 1)(y^2 + y +1). But how do you get there? How do you expand (y^3 + 8) and (y^3 - 1) seperately? Also, what about (27x^4 + 8x)?

I know what the answers are but I just don't know how to get to it. Can somebody help?

Thank you!!

Umm..maybe it's called factoring then..because the answer is supposed to be really long.

Answer 1

Your question can be restated as:
How do I factor cubic polynomials to find the roots?
The roots are the values of y (in this case) that make the equation go to zero.

y^3 + 8 = 0 is a cubic poly equation. It has 3 roots with at least one real root. There are several methods to find roots of polys. They are classified as iterative methods, and analytic methods.

an example of the iterative method is the Lin-Bairstow method. See following link:
http://home.att.net/~srschmitt/script_cubic.html

an example of the analytic method is that of Cardano.
See following link:
http://www3.telus.net/thothworks/Quad3Deg.html

Both links have viewable java source code, and both will find your cubic roots.

Note that a factored term is of the form (y - r) where r is the root. So for a root of -2, the factored term is (y + 2).

Also note that a complex conjugate root pair can always be expressed as a quadratic with real coefficients.
(y-r)(y-r') = y^2 - [2*Real(r)]*y + [Real(r)^2 + Imag(r)^2]
For example r = 1 + i1.732, then (y-r)(y-r') = y^2 - 2y + 4

As for your quartic, it has a root at zero, therefore it is really only a cubic in disguise, (27x)*(x^3 + 8/27).

Answer 2

no you dont expand (y^3 + 8) seperatly.

(a + b) * ( A + B ) = a*(A + B) + b*(A + B); ( fyi this is called the distributive law )

a*(A + B) + b*(A + B) = a*A + a*B + b*A + b*B.


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Ok factoring instead of expanding
Y^3 + 8 = I did this in your other question ...

I will do y^3 - 1

i write y^3 - 1=0

and i see that 1 is a solution of y^3 -1 = 0.
thus y-1 is a factor of y^3-1.
nnow i do a division to find the other factor

y-1 | y^3 - 1 ....... | y^2 + y + 1

........y^3-y^2
.......------------
......... -1 + y^2
..........-y + y^2
..........------------
......... -1 + y
........ -1 + y
==============

so y^3 - 1 = (y-1)(y^2 + y + 1)

next i try to factorise the second term y^2 + y + 1
i dont see it thus i use the dumb abc formula and see that there is no fator in it.

? is it clear ?

Answer 3

I don't know why you think the answer is so complicated!
It is a simple problem. y^3 x y^3 = y^9
y^3 x -1 = -y^3
8 x y^3 = 8y^3
8 x -1 = -8
Add up = y^9-y^3+8y^3-8
Simplify = y^9 =7y^3 -8 QED

Answer 4

ph------ you need more confidence because you are absolutely right. The way the problem is worded to expand would be to foil and your answer is exactly RIGHT and that's why you got my thumbs up!!

y^6 + 7y^3 - 8

Answer 5

(y^3 + 8)(y^3 - 1) = (y^6 - y^3 +8y^3 - 8)
= y^6 + 7y^3 -8
Thats my answer... think its wrong though

Answer 6

Use the FOIL method:
First - Multiply the first term in each set of parentheses
Outer - Multiply the outer term in each set of parentheses
Inner - Multiply the inner term in each set of parentheses
Last - Multiply the last term in each set of parentheses

Answer 7

multiply each element in the first bracket to the others in the other bracket