AP EAMCET201922 Apr 2019Morning ShiftMathematicsQuadratic EquationActual
Let ( (x)= x (x^2+1 )(x+1) ). If (a, b ) and (c ) are the roots of the equation (x^3-3 x+ =0,( 0) ). Then, ( (a) (b) (c)= )
Options
- A( )
- B( - ( +2) ( ^2+16 ) )
- C( ( +2) )
- D( ( +2) ( ^2+16 ) )
Correct answer
D. ( ( +2) ( ^2+16 ) )
Step-by-step solution
( aligned & Given, (x)= x (x^2+1 )(x+1) & (a) (b) (c)= a b c (1+a)(1+b)(1+c) (1+a^2 ) (1+b^2 ) (1+c^2 ) & = a b c (1+a+b+c+a b+b c+c a+a b c) (1+a^2+b^2+c^2 . & . +a^2 b^2+b^2 c^2+c^2 a^2+(a b c)^2 ) aligned ) Also, given that (a, b ) and (c ) are roots of cubic equation ( aligned x^3-3 x+ & =0. a b+b c+c a & =-3 (i) a+b+c & =0 (ii) aligned ) and (a b c=- )...(iii) Squaring Eq. (ii), we get, ((a+b+c)^2=0 ) ( aligned a^2+b^2+c^2+2(a b+b c+c a) & =0 a^2+b^2+c^2 & =6 aligned ) Similarly, (a^2 b^2+b^2 c^2+c^2 a^2=9 ) (