Question

plzzz ans....Thnk u....Second part

Answer 1

since , a : b :: b : c

=> a / b = b / c

=> b² = ac

LHS = 1 / a³ + 1 / b³ + 1 / c³

=> ( b³c³ + a³c³ + a³b³ ) / a³b³c³

=> [ b² × bc³ ] / a³b³c³ + [ ( ac )² × ac ] / a³b³c³ + [ a³b × b² ] / a³b³c³


=> ( ac × bc³ ) / a³b³c³ + [ ( b² )² × ac ] / a³b³c³ + ( a³b × ac ) / a³b³c³


=> c / a²b² + [ ( b⁴ ) × ac ] / a³b³c³ + a / b²c²

=> a / b²c² + b / a²c² + c / a²b²

as , in first eqn , ab in numerator got cancelled out by a³b³ in denominator and c⁴ got cancelled out by c³

in second eqn , ac in numerator got cancelled out by a³c³ in denominator , and b⁴ got cancelled out by b³

in third eqn , bc in denominator got cancelled out by b³c³ in denominator and a⁴ got cancelled out by a³


=> since , LHS = RHS

hence PROVED


HOPE THIS HELPS