Question

solve the equation?
x^6-9x^3+8=0

Answer 1

my q is
x^6-9X^3+8=0

Answer 2

Think of is as Quadratic. x^3 = ???

Answer 3

yeah but how it wil solve

Answer 4

How would you solve this? y^2 - 9y + 8 = 0

Answer 5

Hint: Let x^3 = y, then x^6 = y^2

Answer 6

i did that but i have got only two s.s which are {2,1}
but its solution set should have 6 answers

Answer 7

You quit too soon.

(x^3 - 8)(x^3 - 1) = 0

Factor each of those two factors (difference of cubes) and all six solutions will present themselves.

Answer 8

yes i did
x^3=8 , x^3=1

Answer 9

x^3=2^3 ' x^3=1^0.5

Answer 10

now what dear i have got only two
will you plz help me

Answer 11

That's no good. You didn't FACTOR.

(x^3 - 8) = (x - 2)(x^2 + 2x + 4)

Answer 12

oh gr8 thank u so much

Answer 13

Write it in your brain -- factor, factor, factor, factor, factor....

Answer 14

really thanks tkhunny
mje bht tnsn thi ab bilkul khtam ho gai

Answer 15

another problem
how to solve this
8x^6-19x^3-27=0

Answer 16

\[x^6-9 x^3+8=(x-2) (x-1) \left(x^2+x+1\right) \left(x^2+2 x+4\right) \]Only two real roots, Refer to the attached plot.

Answer 17

@RanaFahdi You DO know what I'm going to say, right? Is it written in your brain?

(x^3 + 1)(8x^3 - 27) = 0

Now what?

Note: I do agree with RobTobey, but keep in mind that the other four are from quadratic factors and can be found easily, even though they are Complex and not Real.

Answer 18

@RanaFahdi
\[8 x^6-19 x^3-27=(x+1) (2 x-3) \left(x^2-x+1\right) \left(4 x^2+6 x+9\right) \]There are only two real roots, -1 and 3/2.
The other four are:\[\pm (-1)^{1/3} \text{and} \frac{3}{4} \left(-1\pm i \sqrt{3}\right) \]
A plot is attached.