Question

if x=1/3-√8 and y=1/3+√8, the value of (x+y)² is (a)32 (b)35 (c)36 (d)38​

Answer 1

Given :-

$x = \dfrac{1}{3 - \sqrt{8} }$

$y = \dfrac{1}{3 + \sqrt{8} }$

To find :-

$(x + y) {}^{2}$

Solution :-

First we find (x + y) value

$x + y = \dfrac{1}{3 - \sqrt{8} } + \dfrac{1}{3 + \sqrt{8} }$

Take LCM to the denominator

$\dfrac{3 + \sqrt{8} + 3 - \sqrt{8} }{(3 + \sqrt{8} )(3 - \sqrt{8}) }$

$\dfrac{6}{(3) {}^{2} - ( \sqrt{8}) {}^{2} }$

Denominator is in form of (a+b)(a-b) = a²-b²

$\dfrac{6}{9 - 8}$

$\dfrac{6}{1}$

$so \: (x + y) = 6$

Now finding (x+y)²

(x + y)² = (6)²

(x + y)² = 36 (option-c)

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Know more some algebraic identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( 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