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# help

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If a^2 - 3a + 9 = 0, then find a^3.

Nov 16, 2019

#1
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The first step in this is to separate $$a^2$$ from the rest of the function.

$$a^2-3a+9=0$$ becomes $$a^2=3a-9$$.

Now there is just one step to find $$a^3$$. Can you take it from here?

Nov 16, 2019
#3
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Withdrawn.

Melody  Nov 16, 2019
edited by Melody  Nov 16, 2019
#4
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I think that this is what zzz had in mind.

.$$\displaystyle \text{Given that }a^{2}-3a+9=0,\dots(1),\\a^{2}=3a-9,\\\text{so}\\a^{3}=3a^{2}-9a.\\ \text{But from (1), }\\a^{2}-3a=-9,\\ \text{so}\\3(a^{2}-3a)=a^{3}=-27.$$

Guest Nov 16, 2019
#5
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ok thanks.

Maybe you are right.

In which case I offer my appologies to ZZZZZZZZZZ.

Sorry ZZZZZZ

Melody  Nov 16, 2019
#2
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a^2  -3a  + 9   = 0   subtract  9 from both sides

a^2  - 3a   =   -9        complete the square on a

a^2  - 3a  + 9/4  =  -9  + 9/4

(a - 3/2)^2  =  -27/4       take both roots

a - 3/2  =  ±√ [-27] / 2

a  =  [3 ± 3i√3 ]

_________

2

When a  = [ 3 + 3i√ 3]

__________   then   a^3  =

2

( [ 3 + 3i√ 3] / 2)^2  ( [3 + 3i√ 3 ] / 2 )  =

[ 9 + 18i√3 + 9i^2 (3) ]

_______________        *   (  [ 3 + 3i√3] / 2)  =

4

[ -18 + 18i√3 ]

_____________     (  [ 3 + 3i√3 ] / 2 )  =

4

[ -9 + 9i√3]            [  3 + 3i√3]

__________  *    __________   =

2                          2

[-27 + 27i√3 + 27i√3 - 27*3 ]         [ -27 * 4 ]

______________________    =  __________   =                -27

4                                          4

The same result is obtained for a^3 when  a  =  ( 3 - 3i√3  )  / 2   Nov 16, 2019