Hello,
Need help in these problems
1-(x+y)^3
(x+y)^3+1
1-a^6x^6
I am having some trouble figuring these out, if you can help me please, thanks.
If you can key it out fully would be great to help me understand.
Thank you
Joanne
Hello,
Need help in these problems
1-(x+y)^3
(x+y)^3+1
1-a^6x^6
I am having some trouble figuring these out, if you can help me please, thanks.
If you can key it out fully would be great to help me understand.
Thank you
Joanne
you have $\displaystyle 1^3 - (x + y)^3$
apply the difference of two cubes formula: $\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
here $\displaystyle a = 1$, $\displaystyle b = x + y$
try it
formula: $\displaystyle a^3 + b^3 = (a + b)(a^2 - ab + b^2)$(x+y)^3+1
here, $\displaystyle a = x + y$, $\displaystyle b = 1$
write it as $\displaystyle 1 - [(ax)^3]^2$.1-a^6x^6
applying the difference of two squares formula, we get: $\displaystyle [1 + (ax)^3][1 - (ax)^3]$
now, apply the sum of two cubes formula to the first factor, and the difference of two cubes formula to the last factor
note the factoring patterns for the sum and difference of two cubes ...
$\displaystyle a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
$\displaystyle a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
for the first problem ...
$\displaystyle a$ is $\displaystyle 1$
$\displaystyle b$ is $\displaystyle (x+y)$
$\displaystyle 1^3 - (x+y)^3 = [1 - (x+y)][1^2 + 1(x+y) + (x+y)^2]$
Answers to first one is
1-(x+y)^3 = (1-x-y)(1+x+y+(x+y)^2 which I got now, thanks.
(x+y)^3+1 = I can't find the full answer to this, sorry.
1-a^6x^6 = (1-ax)(1+ax)(1+ax+a^2x^2)(1-ax+a^2x^2)
This one I am still stuck on doing. I must be doing something wrong, and can't figure it out.
Can someone completely display this answer, thanks.
Joanne
1-a^6x^6 = (1-ax)(1+ax)(1+ax+a^2x^2)(1-ax+a^2x^2)
That is the answer and I am trying this and still not getting anywhere.
I get (1-a^2x^2) (1^2+ax+a^4x^4)
So then you get(1-ax)(1+ax) but stuck on this if it's suppose to look like the answer above.
HELP PLEASE.
thks
Jo
i told you how to get to that answer in my first post. apply the difference of two squares first, then the sum/difference of two cubes.
if you apply the difference of two cubes first (which makes the problem harder) you would get $\displaystyle [1 - (ax)^2][1 + (ax)^2 + (ax)^4]$
you would then need to factor $\displaystyle 1 - (ax)^2$ using the difference of two squares, and to factor $\displaystyle 1 + (ax)^2 + (ax)^4$, you would need to write it as $\displaystyle (ax)^4 + 2(ax)^2 + 1 - (ax)^2 = [(ax)^2 + 1]^2 - (ax)^2$ and use the difference of two sqaures again. See? too much work. review my first post and follow the suggestion there