Question 624702
4(3-x)^4/3 - 5 = 59 
<pre>
{{{matrix(2,1,"",4(3-x)^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

Add 5 to both sides 

    {{{matrix(2,1,"",4(3-x)^(4/3))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",64)}}}

Divide both sides by 4

    {{{matrix(2,1,"",(3-x)^(4/3))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",16)}}}

Now we use the principle of fraction exponents to radicals {{{matrix(2,1,"",A^(B/C))}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",(root(C,A))^B)}}}

    {{{(root(3,3-x))^4}}}{{{""=""}}}{{{16}}}

We use the principle of even roots, in this case 4th roots. When taking
even roots we must use ± on the right side. Now since 16 = 2·2·2·2 = 2<sup>4</sup>,
then the 4th root of 16 is 2. So we have:

    {{{root(3,3-x)}}}{{{""=""}}}{{{"" +- 2}}}

Next to get rid of the cube root we cube both sides:

    {{{(root(3,3-x))^3}}}{{{""=""}}}{{{("" +- 2)^3}}}

    3-x = ±8

If we use the +

    3-x = 8
     -x = 5
      x = -5

If we use the -

    3-x = -8
     -x = -11
      x = 11

So we get two solutions -5, and 11.  We check to see if either one
is extraneous:

Checking x = -5,

{{{matrix(2,1,"",4(3-x)^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

{{{matrix(2,1,"",4(3-(-5))^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

{{{matrix(2,1,"",4(3+5)^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

{{{matrix(2,1,"",4(8)^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

Write 8 as 2³

{{{matrix(2,1,"",4(2^3)^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

Multiply the exponents to remove the parentheses:

4(2<sup>4</sup>) - 5 = 59
4(16) - 5 = 59
64 - 5 - 59
    59 = 59

So -5 is a valid solution.

Checking x = 11:


{{{matrix(2,1,"",4(3-x)^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

{{{matrix(2,1,"",4(3-(11))^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

{{{matrix(2,1,"",4(3-11)^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

{{{matrix(2,1,"",4(-8)^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

Write 8 as 2³

{{{matrix(2,1,"",4((-2)^3)^(4/3))}}}{{{matrix(2,1,"",""-"")}}}{{{matrix(2,1,"",5)}}}{{{matrix(2,1,"",""="")}}}{{{matrix(2,1,"",59)}}}

Multiply the exponents to remove the parentheses:

4((-2)<sup>4</sup>) - 5 = 59
4(16) - 5 = 59
64 - 5 - 59
    59 = 59

So 11 is also a valid solution.

Solutions:  -5 and 11

Edwin</pre>