# Solving a, what am I doing wrong.

• Aug 15th 2012, 12:09 PM
Rpuxster
Solving a, what am I doing wrong.
Can someone please tell me where am I going wrong?
Solve a
b=a^3/\sqrt[3]{a}
b= a^3/1/a^3
b= a^3*1/a^3/1/a^3*1/a^3
b= a^3/1*1/a^3
b=1

This is not right ... don't know where to start or where to go with it.

c = b^-2/\sqrt[3]{a}
c = 1/b^2/*c^1/2
c = 1/b^2*b^2/1*a^1/2
c = b^2*a^1/2
c = b^2*a^1/2/b^2
c/b^2 = a^1/2
a^1/2 = c/b^2

I am going way off the tracks here, please correct me.

• Aug 15th 2012, 12:18 PM
skeeter
Re: Solving a, what am I doing wrong.
$\displaystyle b = \frac{a^3}{\sqrt[3]{a}}$

$\displaystyle b = \frac{a^3}{a^{1/3}}$

$\displaystyle b = a^{8/3}$

$\displaystyle b^{3/8} = a$

$\displaystyle c = \frac{b^{-2}}{\sqrt[3]{a}}$

$\displaystyle \sqrt[3]{a} = \frac{b^{-2}}{c}$

$\displaystyle \sqrt[3]{a} = \frac{1}{b^2c}$

$\displaystyle a = \frac{1}{b^6c^3}$
• Aug 15th 2012, 12:21 PM
Rpuxster
Re: Solving a, what am I doing wrong.
Thank you, I can see where I went wrong ...
• Aug 15th 2012, 12:35 PM
Rpuxster
Re: Solving a, what am I doing wrong.
HallsofIvy, plase tell me what happened in problem one, in step 3 - thank you.
$\displaystyle b = a^{3/8}$
• Aug 15th 2012, 12:44 PM
Rpuxster
Re: Solving a, what am I doing wrong.
... Ok i have it 3*3-1 :D