How to solve (1-a^2)÷(((1-a^(3/2))/(1-a^(1/2))+a^(1/2))*((1+a^(3/2))/(1+a^(1/2))-a^(1/2)))+1, a>=0, a≠1 ?

1 Answer
May 5, 2018

2/(1-a), (a ge0, a!=1).

Explanation:

Let, a^(1/2)=x. :. a^(3/2)=x^3.

:. (1-a^(3/2))/(1-a^(1/2)),

=(1-x^3)/(1-x),

={cancel((1-x))(1+x+x^2)}/cancel((1-x)).

:. (1-a^(3/2))/(1-a^(1/2))=(1+x+x^2).

:. (1-a^(3/2))/(1-a^(1/2))+a^(1/2)=(1+x+x^2)+x=1+2x+x^2, or,

(1-a^(3/2))/(1-a^(1/2))+a^(1/2)=(1+x)^2.......(star^1).

On the similar lines, we can have,

(1+a^(3/2))/(1+a^(1/2))-a^(1/2)=(1-x)^2........(star^2).

Combining (star^1) and (star^2), we have,

{((1-a^(3/2))/(1-a^(1/2))+a^(1/2))( (1+a^(3/2))/(1+a^(1/2))-a^(1/2))}=(1-x^2)^2.

Hence, the Exp. =(1-x^4)-:(1-x^2)^2+1,

={(1+x^2)(1-x^2)}-:(1-x^2)^2+1,

=(1+x^2)/(1-x^2)+1,

={(1+x^2)+(1-x^2)}/(1-x^2),

=2/(1-x^2),

=2/(1-a), (a ge 0, a!=1).