I'm having troubles solving the following problem.

x^6 + 5x^3 -24 = 0

Mouseman

2. Originally Posted by Mouseman
I'm having troubles solving the following problem.

x^6 + 5x^3 -24 = 0

Mouseman
(x^3 + 8)(x^3 - 3) = 0

x = cubert(-8) or x = cubert(3)

3. $\displaystyle \sqrt[3]{-8}=-2$

4. Originally Posted by Mouseman
I'm having troubles solving the following problem.

x^6 + 5x^3 -24 = 0

Mouseman
Rewrite this as

$\displaystyle (x^3)^2+5x^3-24=0$

Now let $\displaystyle u=x^3$

giving us

$\displaystyle u^2+5u-24=0$

which implies that

$\displaystyle u=\frac{-5\pm\sqrt{25-4(-24)(1)}}{2}=\frac{-5\pm{11}}{2}$

So either

$\displaystyle u=-8$

or

$\displaystyle u=3$

So backsubbing we get

$\displaystyle x^3=-8\Rightarrow{x=-2}$

or

$\displaystyle x^3=3\Rightarrow{x=\sqrt[3]{3}}$

Note this is very useful trick...for example say you were given

solve

$\displaystyle \sin^2(x)+2\sin(3x)-8=0$

Let $\displaystyle u=\sin(x)$

do the same thign

or

$\displaystyle e^{2x}+10e^x-11=0$

rewriting this as

$\displaystyle (e^x)^2+10e^x-11=0$

let $\displaystyle e^x=u$

same trick

Remember that when you solve the u formula with the quadratic formula that you must take the inverse to get the actual answer

for example

in the sin(x) one

lets say after you sub you get

$\displaystyle u=5$

then by backsubbing you get

$\displaystyle \sin(x)=5\Rightarrow{x=arcsin(5)}$

This is a very useful tool