Question

If x + y = -4, find the value of(I shall type in equation editor).

Answer 1

\[x ^{3} + y ^{3} - 12xy + 64\]

Answer 2

Concept involved: algebraic identities. I have racking my brains for the past half an hour. Someone please help.

Answer 3

answer is zero

Answer 4

Please explain...

Answer 5

and the reason it is zero is because this factors as
\[{(x+y+4) (x^2-x y-4 x+y^2-4 y+16)}\]

Answer 6

Did you find it using identities?

Answer 7

so if \(x+y=-4\) then the first term is 0

Answer 8

OK, I get it but I don't understand how you factored to get whatever you got.

Answer 9

Please explain the factoring part ...

Answer 10

give me a minute

Answer 11

ok ready?

Answer 12

we want to make it look like \(a^3+b^3\) where \(a=x+y, b=4\) because this almost looks like
\[(x+y)^3+64\] which you can factor as the sum of two cubes

Answer 13

What do you do with -12xy?

Answer 14

so \[(x+y)^3=x^3+y^3+3x^3y+3xy^2+y^3\] so if you replace \(x^3+y^3\) by \((x+y)^3\) you have to subtract off \(3x^2y+3xy^2\) and you get
\[x ^{3} + y ^{3} - 12xy + 64=(x+y)^2+64-3x^2y-3xy^2-12xy\]

Answer 15

let me know if that last step was clear. by replacing
\[x^3+y^3\] by
\[(x+y)^3\] i added \[3x^2y+3xy^2\] so i had to subtract it off

Answer 16

Yeah I got that.

Answer 17

i also made a typo it should have been
\[x ^{3} + y ^{3} - 12xy + 64=(x+y)^3+64-3x^2y-3xy^2-12xy\]

Answer 18

now you can factor the second part, the part after the 64 as
\[-3x^2y-3xy^2-12xy=-3xy(x+y+4)\] easily

Answer 19

ok

Answer 20

now we can factor the whole thing, first the sum of two cubes
\[a^3+b^3=(a+b)(a^2-ab+b^2)\] with
\[a=x+y, b=4\] and it will work out

Answer 21

you get
\[(x+y)^3+64-3xy(x+y+4)\]
\[=(x+y+4)((x+y)^2-4(x+y)+16)-3xy(x+y+4)\] giving you a common factor of
\[x+y+4\] in both tersm

Answer 22

yes and then factor the first part as the sum of two cubes as i wrote above. you see both terms have a common factor of \(x+y+4\)

Answer 23

so you can rewrite as
\[(x+y+4)((x+y)^2-4(x+y)+16-3xy)\] and you can do some algebra to clean up the second factor, but it is unnecessary for this problem since you already have the factor \(x+y+4\) that you want

Answer 24

adding and subtracting to factor is a good trick to know, used also for completing the square.
sometimes called "sophie germain" trick

Answer 25

it is a bunch of algebra, but i think the steps i wrote are right. if it is unclear let me know

Answer 26

Thank you so much!! ... I got it ... got to go now!

Answer 27

it is also unclear how you were supposed to know this
yw