Question 944238
<pre>
The other tutor's answer above is incorrect.  The other tutor gave the
solution for {{{ (16x)^(-3) = (2x)^(-6) }}}.  The 2 and the 16 are not in
parentheses so they are not raised to those negative powers.  Only x is
raised to those negative powers in your equation.

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{{{ 16x^(-3) = 2x^(-6) }}}

Divide both sides by 2

{{{ 8x^(-3) = x^(-6) }}}

Write the factors with negative exponents as denominator
factors with positive exponents:

{{{ 8/x^3 = 1/x^6 }}}

Cross-multiply:

{{{8x^6=x^3}}}

Get 0 on the right sides:

{{{8x^6-x^3=0}}}

Factor out x³

{{{x^3(8x^3-1)=0}}}

{{{x^3=0}}},  {{{8x^3-1=0}}}

{{{x=0}}},    {{{(2x)^3-1^3=0}}}

0 is not a solution because 0 may only be raised to 
positive powers, and the original has negative powers.
The second equation requires the factorization of the
difference of two cubes:


              {{{(2x-1)((2x)^2+2x*1+1^2)}}}

              {{{(2x-1)(4x^2+2x+1)}}}

              {{{2x-1=0}}}   {{{4x^2+2x+1=0}}}

              {{{2x=1}}}     {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
             
              {{{x=1/2}}}    {{{x = (-2 +- sqrt( 2^2-4*4*1 ))/(2*4) }}}

                             {{{x = (-2 +- sqrt(4-16 ))/8 }}}

                             {{{x = (-2 +- sqrt(-12 ))/8 }}}

                             {{{x = (-2 +- i*sqrt(12 ))/8 }}}

                             {{{x = (-2 +- i*sqrt(4*3 ))/8 }}}

                             {{{x = (-2 +- 2i*sqrt(3 ))/8 }}}

                             {{{x = (2(-1 +- i*sqrt(3 )))/8 }}}

                             {{{x = (-1 +- i*sqrt(3))/4}}}

So there are three solutions {{{1/2}}}, {{{(-1 + i*sqrt(3))/4}}}, {{{(-1 - i*sqrt(3))/4}}}  


Edwin</pre>