Question

Solve \frac{{(\sqrt{3}+1)}^{2}}{{(\sqrt{3}-1)}^{2}}+\frac{{(\sqrt{3}-1)}^{2}}{{(\sqrt{3}+1)}^{2}}+1(3​−1)2(3​+1)2​+(3​+1)2(3​−1)2​+1

Answer 1

Step-by-step explanation:

I will tell you what formulae I used.

${(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

${(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab$

$(a + b)(a - b) = {a}^{2} - {b}^{2}$

${(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

If you like the process please comment.

Answer 2

Given to find the value of :-

$\dfrac{ (\sqrt{3} + 1) {}^{2} }{( \sqrt{3} - 1) {}^{2} } + \dfrac{( \sqrt{3 } - 1) {}^{2} }{( \sqrt{3} + 1) {}^{2} } + 1$

$\dfrac{ (\sqrt{3} + 1) {}^{2} }{( \sqrt{3} - 1) {}^{2} } + \dfrac{( \sqrt{3 } - 1) {}^{2} }{( \sqrt{3} + 1) {}^{2} } + 1$

SOLUTION :-

Simplify the numerator and denominator which is in form of (a+b)² and (a-b)²

• (a+b)² = a² + 2ab + b²
• (a-b)² = a² -2ab +b²

First we simplify the first term $\dfrac{( \sqrt{3}) {}^{2} + (1) {}^{2} + 2(1)( \sqrt{3} ) }{( \sqrt{3}) {}^{2} + (1) {}^{2} - 2(1)( \sqrt{3}) }$

$\dfrac{3 + 1 + 2 \sqrt{3} }{3 + 1 - 2 \sqrt{3} }$

$\dfrac{4 + 2 \sqrt{3} }{4 - 2 \sqrt{3} }$

So,

$\dfrac{ (\sqrt{3} + 1) {}^{2} }{( \sqrt{3} - 1) {}^{2} } = \dfrac{4 + 2 \sqrt{3} }{4 - 2 \sqrt{3} }$

Now , we shall simplify the 2nd term

$\dfrac{( \sqrt{3} - 1) {}^{2} }{( \sqrt{3} + 1) {}^{2} }$

$\dfrac{( \sqrt{3}) {}^{2} + (1) {}^{2} - 2(1)( \sqrt{3} ) }{( \sqrt{3}) {}^{2} + (1) {}^{2} +2(1)( \sqrt{3}) }$

$\dfrac{4 - 2 \sqrt{3} }{4 + 2 \sqrt{3} }$

So,

$\dfrac{( \sqrt{3} - 1) {}^{2} }{( \sqrt{3} + 1) {}^{2} } = \dfrac{4 - 2 \sqrt{3} }{4 + 2 \sqrt{3} }$

$\dfrac{ (\sqrt{3} + 1) {}^{2} }{( \sqrt{3} - 1) {}^{2} } + \dfrac{( \sqrt{3 } - 1) {}^{2} }{( \sqrt{3} + 1) {}^{2} } + 1$

$\dfrac{4 + 2 \sqrt{3} }{4 - 2 \sqrt{3} } + \dfrac{4 - 2 \sqrt{3} }{4 + 2 \sqrt{3} } + 1$

$\dfrac{(4 + 2 \sqrt{3}) {}^{2} + (4 - 2 \sqrt{3}) {}^{2} + (4 - 2 \sqrt{3})(4 + 2 \sqrt{3} ) }{(4 - 2 \sqrt{3})(4 + 2 \sqrt{3} )}$

$\dfrac{2(16 + 12) + 16 - 12}{16 - 12}$

$\dfrac{2(28) + 4}{4}$

$\dfrac{56 + 4}{4}$

$\dfrac{60}{4}$

$= 15$

Used formulae:-

• (a+b)² = a² + 2ab + b²
• (a-b)² = a² -2ab + b²
• (a+b)(a-b) = a²-b²
• (a+b)² +(a-b)² = 2a² + 2b²

Know more some of the formulae

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc