Question

Solve the following using substitution.

Answer 1

\[(x+1)^{3}=(x-1)^{3}+26\]

Answer 2

substitute with....?

Answer 3

by any variable k

Answer 4

\[k^3=(k-2)^3+26\]

Answer 5

okayy. but how would you solve for k?

Answer 6

\[k^3=k^3-6k^2+12k-8+26\]

Answer 7

how do you multiply out cubes??

Answer 8

i got \[k ^{3}-8\]

Answer 9

\[(x-y)^3=x^3+3x^2y-3xy^2-y^3\]

you should get because k^3 cancel \[6k^2-12k-18=0\]

Answer 10

okay. thanks!

Answer 11

last question...

Answer 12

okay

Answer 13

how would you factor \[(x+y)^{3}\]

Answer 14

everything is positive but the same as the one with - except for it is all +

Answer 15

lets verify
\[(x+y)^3=(x+y)(x+y)(x+y)\]
open two brackets

Answer 16

THANKS!!!

Answer 17

\[x^3+y^3=...\]how can you factcor that

Answer 18

good but without the 2xy jus xy

Answer 19

oops

Answer 20

thats correct for \[(x+y)^3\]not for\[x^3+y^3\]

Answer 21

okay...then what is it for x3+y3

Answer 22

we use \[(x+y)^3=x^3+3xy^2+3x^2y+y^3\]
to get it

Answer 23

???

Answer 24

on the equation do you see any \[x^3+y^3\]

Answer 25

yes...

Answer 26

can you take everything to the other side except for that

Answer 27

\[x^3+y^3=(x+y^3)-3xy^2-3x^2y\]

Answer 28

furthermore 3xy factorised give
\[(x+y)^3-3xy(x+y)\]

Answer 29

factorize x+y

Answer 30

okayy. but how does that help?

Answer 31

\[(x+y)(x ^{2}-xy+y ^{2})\]

Answer 32

nice thats how i wanted us to factorize

Answer 33

so (x2−xy+y2) would equal 3xy?

Answer 34

no not really,we get that when we say
\[(x+y)[(x+y)^2-3xy]=(x+y)(x^2+2xy+y^2-3xy)\]
\[(x+y)(x^2-xy+y^2)\]

Answer 35

so now we know how factor 3 different cubes

Answer 36

ohh. okay. thanks :)