Question

find the fourth root of 28-16 √3 without using log tables. But use short method. I need it urgently.

Answer 1

Write $28-16\sqrt{3}$ as $(1)^{4}-4(1)^{3}(\sqrt{3})+6(1)^{2}(\sqrt{3})^{2}-4(1)(\sqrt{3})^{3}+(\sqrt{3})^{4}$.
This is expansion of $(1-\sqrt{3})^{4}$.
Hence, answer is $(1-\sqrt{3})$.

Answer 2

     (x - y√3)^4 =  28 - 16 √3

x⁴ - 4 x³ y√3 + 6 x² * 3 y² - 4 x y³ 3√3 + 9 y⁴  = 28 - 16 √3

   x⁴ + 18 x²y² + 9 y⁴ = 28  -- equation 1
      x² y + 3 x y² = 4  -- equation 2

Solve them, if you can.  Or follow the method below.  Perhaps there is a simpler method too.

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I do that in two smaller steps .  if  i cannot do more complex way.

(x - y√3)² = 28 - 16 √3

Hence,   x² + 3 y² = 28          and  2 x y = 16  =>  y = 8 / x
            x² + 3 * (8/x)² = 28

         x⁴ - 28 x² + 192 = 0   
x² = [ 28 + - √(28² - 4*192) ] / 2  = 16 or 12      Hence  x = +4 or -4  or √12  or -√12
y = +2,  - 2,  +4/√3 , - 4/√3.
   You can factorise above as (x² - 16) (x² -12 ) = 0

Now  square root of  28 - 16 √3 =  4 - 2 √3 .  We can take  negative values also.  But then their square roots will be imaginary.  You can also that do if required.
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Now  find the square root of  4 - 2 √3  like above.

(x - y√3)² = 4 - 2 √3
solving  like above
      
   x² = 3   or  1

so  x = -1, + 1,  + √3  or  -√3.

Hence,    answer  is   1- √3    or  √3 - 1