Question

x+y=a and xy=b then the value of 1/x^3+1/y^3

Answer 1

1/x^3+1/y^3
=(y^3+x^3)/x^3*y^3
=(x+y)(x^2-xy+y^2)/(xy)^3
= a[(x+y)^2-2*xy-xy]/(xy)^3
= a[a^2-3xy]/b^3
= (a^3-3ab)/b^3

Answer 2

Hi ,

x + y = a -----( 1 )

xy = b ------( 2 )

do the cube of equation ( 1 )

( x + y )³ = a³

x³ + y³ + 3xy( x + y ) = a³

x³ + y³ + 3ba = a³ [ from ( 1 ) and ( 2 ) ]

x³ + y³ = a³ - 3ab ----( 2 )

according to the problem given,

1/x³ + 1/y³

= ( x³ + y³ )/x³y³

= ( x³ + y³ ) / ( xy )³

= (a³ - 3ab ) / b³ [ from ( 3 ) and ( 2 ) ]

I hope this helps you.

:)