Question

If x+y=9 and xy=20 find the value of x^3-y^3
Class 9 question

Answer 1

(x+y)^3=x^3+y^3 +3xy (x+y)
(9)^3= x^3+y^3 +3×20(9)
729=x^3+y^3+540(after solving it)
x^3 +y^3=729-540
x^3+y^3=189

Answer 2

x + y = 9 --eq(1) , xy = 20

on squaring both side(eq.1) , we get

(x + y)^2 = (9)^2

x^2 + y^2 + 2xy = 81

x^2 + y^2 + 2(20) = 81

x^2 + y^2 = 81 - 40 = 41

x^2 + y^2 = 41. ______√_________

x^2 + y^2 - 2xy + 2xy = 41

(x - y)^2 + 2xy = 41

( x - y )^2 + 2(20) = 41

( x - y )^2 = 41 - 40

( x - y )^2 = 1 => x - y = 1 __√_______

x^3 - y^3 = ( x - y ) (x^2 + xy+ y^2 )

=1 ( x^2 + 20 + y^2 )

=( x^2 + y^2 + 20 )

=41 + 20 = 61

therefore, x^3 - y^3 = 61
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Your answer : 61
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