Question

if a=2-√5/2+√5 and b=2+√5/2-√5 find (a+b)³

Answer 1

Hi friend, Harish here.

Here is your answer;

Given that,

$a = \frac{2- \sqrt{5} }{2+ \sqrt{5}} : b = \frac{2+ \sqrt{5} }{2- \sqrt{5}}$

To find,

The value of ( a+b )³.

Solution:
 
First let us rationalize both, a & b.

⇒ $a = \frac{2- \sqrt{5} }{2+ \sqrt{5} } \times \frac{2- \sqrt{5} }{2- \sqrt{5}} = \frac{(2- \sqrt{5})^{2}}{(2+ \sqrt{5})(2- \sqrt{5})}$

We know that, (a+b)(a-b) = a² - b²

Then, ( 2 + √5) ( 2 - √5) = 2² - (√5)² = 4 - 5 = -1

Then,

⇒ $\frac{(2- \sqrt{5})^{2}}{(2+ \sqrt{5})(2- \sqrt{5})} = \frac{4+5-4 \sqrt{5} }{-1} = -( \frac{9-4 \sqrt{5} }{1}) = (4 \sqrt{5} - 9)$

So, a = 4√5 - 9

Now,

⇒ $b =\frac{2+ \sqrt{5} }{2- \sqrt{5} } \times \frac{2+\sqrt{5} }{2+ \sqrt{5}} = \frac{(2+ \sqrt{5})^{2}}{(2+ \sqrt{5})(2- \sqrt{5})}$

⇒  $\frac{4+5+4 \sqrt{5} }{-1}= -( \frac{9+4 \sqrt{5}}{1}) = -9 - 4 \sqrt{5}$

⇒ b = -9 - 4√5

Now,

$( a + b)^{3} = ( -9 + 4 \sqrt{5} - 9 -4 \sqrt{5})^{3} = (-18)^{3} = -5832$

$\bold{Therefore\ (a+b)^{3}\ is\ -5832}$
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Hope my answer is helpful to you.

Answer 2

refer this pic for the answer

if a=2-√5/2+√5 and b=2+√5/2-√5 find (a+b)³

=-5832