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.

Thanks in advance - Roux

Re: Solving a, what am I doing wrong.

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

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

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

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

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

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

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

$\displaystyle a = \frac{1}{b^6c^3}$

Re: Solving a, what am I doing wrong.

Thank you, I can see where I went wrong ...

Re: Solving a, what am I doing wrong.

HallsofIvy, plase tell me what happened in problem one, in step 3 - thank you.

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

Re: Solving a, what am I doing wrong.

... Ok i have it 3*3-1 :D