`(sin^(-1) x)^(3) + (cos^(-1) x)^(3) = a pi^(3)`
`rArr (sin^(-1) x + cos^(-1) x) ((sin^(-1) x + cos^(-1)x)^(2) - 3 sin^(-1) x cos^(-1) x) = a pi^(3)`
`rArr (pi^(2))/(4) - 3 sin^(-1) x cos^(-1) x = 2 a pi^(2)`
`rArr sin^(-1) x ((pi)/(2) - sin^(-1) x) = (pi^(2))/(12) (1 -8a)`
`rArr (sin^(-1) x)^(2) - (pi)/(2) sin^(-1) x = -(pi^(2))/(12) (1 - 8a)`
`rArr (sin^(-1) x - (pi)/(4))^(2) = (pi^(2))/(12) (8a - 1) + (pi^(2))/(16)`
`= (pi^(2))/(48) (32a -1)`
Now, `sin^(-1) x in [-(pi)/(2), (pi)/(2)]`
`rArr - (3 pi)/(4) le sin^(-1) x - (pi)/(4) le (pi)/(4)`
`rArr 0 le (sin^(-1) x - (pi)/(4))^(2) le (9pi^(2))/(16)`
`rArr 0 le (pi^(2))/(48) (32a - 1) le (9pi^(2))/(16)`
`rArr 0 le 32 a - 1 le 27`
`rArr (1)/(32) le a le (7)/(8)`
Thus, the required set of value of `a` is `[(1)/(32), (7)/(8)]`
