Question

A^2+√3a-5 =0 then find the value of( a-√3)^3+1\(a-√3)^3

Answer 1

Answer:

3a - 3/a - 3 = 0

=> 3a² -3 -3a = 0

=> a = {3 +- √(9+36)}/6

=> a= (3+-3√5)/6

=> a = (1+-√5)/2 ………. (1)

So, 1/a = 2/(1+-√5)

We, rationalize the denominator first by taking 1+√5 & then by 1-√5 . In both the cases .. value of (a-1/a) will be the same

2/(1+√5) = 2(1-√5) / (1+√5)(1-√5)

= 2(1-√5) / -4

=> 1/a = (1-√5) / -2 …………….. (2)

By (1) & (2)

(a- 1/a)

= (1+√5)/2 - (1-√5)/-2

= {(1+√5 + ( 1 -√5)} /2

= 2/2 = 1

=> ( a- 1/a) = 1 ………. (3)

Now, using identity

x^3 - y^3 = (x-y)^3 +3 xy ( x-y)

We get, a^3 - 1/a^3 + 2

= ( a- 1/a)^3 + 3(a- 1/a) + 2

By putting the value of (a- 1/a) in the above expression , by …(3)

we get, 1 + 3 + 2

= 6

Step-by-step explanation:

I hope it helps you