Question

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Answer 1

Here is Ur Ans.

x³ + y³ = 126 ... Equation i

x²y + xy² = 30

xy ( x + y ) = 30

( x + y ) = 30/xy ... Equation ii

From Equation i we have

x³ + y³ = ( x + y )³ -3xy ( x + y )

( x + y )³ - 3xy( x + y ) = 126

( x + y )³ - 3 xy ( 30/xy ) = 126

( x + y )³ - 90 = 126

( x + y )³ = 126 + 90

( x + y )³ = 216

x + y = (216)⅓

x + y = 6 ... Equation iii

Now, substitute value of x + y in Equation ii we have

6 = 30/xy

xy = 5 ... Equation iv

y = 5/x

Put value of y in Equation iii we have

x + 5/x = 6

x² + 5 = 6x

x² - 6x + 5 = 0

x² -5x -1x + 5 = 0

x ( x - 5 )-1 ( x -5 ) = 0

( x - 1 ) = 0 OR ( x - 5 ) = 0

x = 1 OR x = 5

For x = 1 y = 5

OR

For x = 5 y = 1

So, ( x , y ) = ( 1 , 5 ) or ( 5 , 1 )