EAMCET Engineering 2010 Solved Paper

Equation x3−ax2+ax−1=0 and α≠1, a≠−1.
we put, (x=‌

1

y

) in given equation
Then, (‌

1

y

)3−a(‌

1

y

)2+a(‌

1

y

)−1=0
‌

1

y3

−‌

a2

y2

+‌

a

y

−1=0
1−a2y+ay2−y3‌=0
⇒‌‌y3−ay2+a2y−1‌=0
Since, the reduced equation is same as original equation by replacing (x=‌

1

y

) ie, reciprocal root of the given equation.
Hence, (x=‌

1

α

) is a root of the given equation