Question

Please answer the following
#3

Answer 1

$\sqrt[2010]{(2 \sqrt{7}-3 \sqrt{3}) \sqrt{55+12 \sqrt{21} } }$
Let us start with √(55+12√21).
55+12√21
=55+2×6×√21
=55+2×2√7×3√3
=28+27+2×2√7×3√3
=(2√7)²+2×2√7×3√3+(3√3)²
=(2√7+3√3)²
∴,√(55+12√21)=2√7+3√3 (neglecting the negative sign)
 $\sqrt[2010]{(2 \sqrt{7}-3 \sqrt{3}) \sqrt{55+12 \sqrt{21} } }$
=$\sqrt[2010]{(2 \sqrt{7}-3 \sqrt{3})(2 \sqrt{7}+3 \sqrt{3}) }$
= $\sqrt[2010]{(2 \sqrt{7})^{2}- (3 \sqrt{3} )^{2} }$
=$\sqrt[2010]{(28-27)}$
=$\sqrt[2010]{1}$
=1
∴, Answer: (2) 1

Answer 2

I will explain two ways of finding the square root here:

we can find the square root in this way:
$\sqrt{55+ 12 \sqrt{21}}=\sqrt{a}+\sqrt{b}\\\\on\ squaring:\\\\55=a+b,\ \ \ 2\sqrt{ab}=2*6\sqrt{21},\\\\ \ \ ab=6^2*21=3*7*2*3*2*3$

a+b = 55      ab = 21*36
you can find  (a-b) by (a-b)² = (a+b)² - 4 ab = 1
so  a = (55+1)/2 = 28     and   b = 27


Now:
$\sqrt{a}=2\sqrt{7},\ \ \sqrt{b}=3\sqrt{3}\\\\Answer=\sqrt[2010] {(2\sqrt7-3\sqrt3)(2\sqrt7+3\sqrt3)}=\sqrt[2010]{28-27}=1$
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Finding a, b  when  a+b = 55   and  ab = 2 * 2 * 7 * 3 * 3 * 3 
find factors of a*b  so that their sum is  55....   they are 28 and 27...