Need help! X^3 question

[ryan_kidz]

If x^2= x+3, then x^3=

sounds simple, but i can't solve it ^-^

thnx for ur help.

 

[topdawgizalaw]

i'll help u if u help me, my thread is titled, urgent need help

 

[Unco]

$\displaystyle x^3 = x \cdot x^2$

 

[Denis]

x^3 = x^1000 * x^(-997) :roll:

 

[stapel]

If x² = x + 3, then:

. . . . .x² - x - 3 = 0

. . . . .x = [1 ± sqrt(13)] / 2

. . . . .x³ = [1 ± sqrt(13)]³ / 8

. . . . .x³ = [1 ± 3sqrt(13) + 39 ± 13sqrt(13)] / 8

. . . . .x³ = [40 + 16sqrt(13)] / 8 or [40 - 16sqrt(13)] / 8

. . . . .x³ = 5 ± sqrt(13)

I'm fairly certain this is not what they're looking for -- they're expecting a nicer answer using cleverness, such as is suggested by Unco's hint -- but I'm not coming up with anything better than the above. Sorry.

Eliz.

 

[ryan_kidz]

thnx for ur helps stapel and the others, i really appreciate it.

i know that is kinda tricky question. but i still can't figure it out.
do u guys want to help me again about this problem?

The key answer is 4x+3

 

[Denis]

If x^2= x+3, then x^3=
thnx for ur help.

x*x = x + 3
x*x*x = x^3
x^3 = (x + 3)*x = x^2 + 3x

 

[ryan_kidz]

yeah.. that's make a sense, but The key answer is 4x+3

confusing huh?

 

[soroban]

Hello, ryan_kidz!

I looked for a clever way to answer this one, but failed . . .
. . I had to resort to Brute Force.

If $\displaystyle x^2\,=\,x\,+\,3$, then \(\displaystyle x^3\,=\.?\)

We have the quadratic: .$\displaystyle x^2\,-\,x\,-\,3 \:= \:0$

Using the Quadratic Formula: .$\displaystyle \displaystyle{x \:= \;\frac{1\, \pm\, \sqrt{13}}{2}}$

Then: . $\displaystyle x \:= \:\left(\frac{1\,\pm\,\sqrt{13}}{2}\right)^3 \;= \;\frac{1\,\pm\,3\sqrt{13}\,+\,39\,\pm\,13\sqrt{13}}{8} \;= \;\frac{40\,\pm\,16\sqrt{13}}{8} \;= \;5\,\pm\,2\sqrt{13}$

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Blast! . . . I just saw the follow-up postings.
. . What a stupid problem!
. . There are dozens of correct answers . . .

I had that answer, too, but I thought it was too silly.


We have: .$\displaystyle x^2 \:= \:x\,+\,3$

Multiply by $\displaystyle x:\;\;x^3 \:= \:x^2\,+\,3x$

Since $\displaystyle x^2 \:= \:x\,+\,3$, we have:

. . . \(\displaystyle x^3 \:= \x\,+\,3)\,+\,3x \:= \:4x\,+\,3\)

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Here are some <u>other</u> correct answers . . .

Since $\displaystyle x^2 \:= \:x\,+\,3$, raise both sides to the power $\displaystyle \frac{3}{2}:$
. . . \(\displaystyle x^3 \:= \x + 3)^{\frac{3}{2}}\)


Since $\displaystyle x \:= \:x^2\,-\,3$, cube both sides: .\(\displaystyle x^3 \:= \x^2\,-\,3)^3\)


We have: .$\displaystyle x^2 \:= \:x\,+\,3$. .Divide both sides by $\displaystyle x:\;\;x \:= \:\frac{x\,+\,3}{x}$
. . Now cube both sides: .$\displaystyle x^3 \:= \:\left(\frac{x\,+\,3}{x}\right)^3$

 

[ryan_kidz]

aah.. haha
yeah u right soroban!

thnx alot of the answers guys!