Question

Given that ‘a’ is a root of the equation x2-x-3=0. Evaluate the value of a3+1/a5-a4-a3+a2​

Answer 1

Given :  ‘a’ is a root of the equation  x² - x  - 3 = 0

To find :   (a³  + 1) / (a⁵ - a⁴ - a³ + a²)

Solution:

x² - x  - 3 = 0

a is a root

=> a² - a  - 3  = 0

=>  a² - a = 3

=> a(a - 1) = 3

(a³  + 1)/(a⁵ - a⁴ - a³ + a²)

= (a³  + 1) /a²(a³ - a² - a  + 1)

= (a³  + 1) /a²(a²(a  - 1) -1(a - 1))

using x³ + y³ = (x + y)(x²- xy + y² )

=  (a + 1) (a² - a  + 1) / a² (a² - 1)(a - 1)

=  (a + 1) (a² - a  + 1) / a² (a + 1)(a - 1)(a - 1)

= (a² - a  + 1) / a²(a - 1)(a - 1)

= (a² - a  + 1)  / (a(a - 1))²

= (a² - a  + 1)  / (a² - a)²

using   a² - a = 3  

=  ( 3 + 1)/3²

= 4/9

(a³  + 1)/(a⁵ - a⁴ - a³ + a²) = 4/9

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