Question

factoring
x^6-26x^3-27=

Answer 1

ok put x^3 = t

Answer 2

then the equation become

t^2 - 26 t -27

Answer 3

now when we factorise them

t^2 + 27t -t -27 = 0

t(t+27) -1(t+27)=0

(t-1)(t+27)=0


: )

Answer 4

now t = -27, 1

t = x^3

=> x^3 = -27
x = -3

and similarly

t = 1

=> x =1

Answer 5

t^2 + 27t -t -27 = 0
it is not t^2-26t-t-27

Answer 6

it not make sense

Answer 7

it can be t^2-27t+t-27

Answer 8

is it?

Answer 9

yea sure it can be the u sttd above ....sry i ..got it wrong ...but my method is right ..so u follow tht

Answer 10

did u get it ...the main step here is x^3=t one ...

Answer 11

ok dear

Answer 12

now frm the right equation u will get

t = 27 ,-1

Answer 13

then when u put t= x^3 ...u will get the answers ..

Answer 14

so it is (t-1)(t-27)=o

Answer 15

no ...
rmmbr the eq

t^2 - 27t +t -27 =0

Answer 16

yes

Answer 17

now

t(t - 27 ) +1 (t - 27) =0

Answer 18

this gives u

(t +1 )(t - 27) = 0

Answer 19

which gives t = -1 , 27

Answer 20

ok

Answer 21

now u solve it for x

u know t = x^3

i'm doing one ...root .. u tell me nther one

x^3 = 27

x = ((3)^3)^1/3
x=3

Answer 22

so it is (x^3+1)(x^3+3^3)

Answer 23

op x^3-3^3

Answer 24

yea sry typo done by me .... u r right

Answer 25

so you have formular a^3 + b^3

Answer 26

and a^3-b^3

Answer 27

but here u dont need it ... wt u can do is

(x^3 + 1)(x^3 -3^3) = 0

the u divide both sides by (x^3 - 3^3)


then u get

(x^3 + 1 ) = 0 . 1 /(x^3 -3^3)

here we know nethng multiplied by zero is zero

so x^3 = -1

x = -1


u get one root by this ...solve the sam way fr another one

Answer 28

though (a^3 - b^3) = (a-b)(a^2 + b^2 +ab)

Answer 29

yes

Answer 30

u can use tht one formula too a^3 - b^3 but it will make it unnecessary lengthy

Answer 31

its result is (x+1)x^2-x+1)(x-3)(x^2+3x+9)

Answer 32

yea thts another way of saying thts all i thought u have to solve fr x .....i didn't u were to make factors ...yea ur solution is CORRECT !

Answer 33

thank for your help too

Answer 34

: )