(3x-1)^3^ = exponents, so it's (3x-1) cubed

Wednesday, July 31, 2013

You did not specify what is the required
information.

I assume that you either need to expand or
find the roots.

First, let us expand between brackets using
the formula:

We know
that:

$\displaystyle{{\left({a}-{b}\right)}}^{{3}}={{a}}^{{3}}-{3}{{a}}^{{2}}{b}+{3}{a}{{b}}^{{2}}-$
b^3

$\displaystyle=\Rightarrow{{\left({3}{x}-{1}\right)}}^{{3}}={27}{{x}}^{{3}}-{27}{{x}}^{{2}}+{9}{x}$
-1

If you need to determine the roots (
zeros):

==> $\displaystyle{{\left({3}{x}-{1}\right)}}^{{3}}=$
0

$\displaystyle=\Rightarrow{3}{x}-{1}=$
0

$\displaystyle=\Rightarrow{3}{x}=$
1

$\displaystyle=\Rightarrow{x}=$
1/3

Help Notes