Question

a³+1_a³+2 if 2a-2_a+1=0​

Answer 1

Answer:

2a - 2/a = (-1)

2[ a-1/a] = (-1)

[a-1/a] = (-1/2)

squaring both sides we get,

(a²+1/a² - 2) = 1/4

Adding both sides 4 we get,

(a² + 1/a² + 2) = 17/4

Square root both sides we get ,

(a+1/a) = √17/2 ---------------(1)

Now, we have to find = [ a³+1/a³ +2 ]

Cubing both sides in equation (1) we get,

(a³ + 1/a³ + 3√17/2) = 17√17/8

(a³ + 1/a³) = (17√17/8) - (3√17/2)

(a³ + 1/a³) = (17√17 - 12√17) /8

(a³ + 1/a³) = 5√17/8

so,

[ a³+1/a³ +2 ] = ( 5√17/8) + 2

[ a³+1/a³ +2 ] = [ (5√17) + 16 ] /8 [Ans.]

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