Question

if (x^2+1)(y^2+1)+16=8(x+y) , then find x^3+y^3​

Answer 1

Answer:

x³ + y³ =  52

Step-by-step explanation:

if (x^2+1)(y^2+1)+16=8(x+y) , then find x^3+y^3​

(x² + 1)(y² + 1) + 16 = 8(x+y)

=> x²y² + 1  + x² + y² + 16 = 8(x + y)

=> x²y² + 1 - 2xy + 2xy + x² + y² + 16 -8(x + y) = 0

=> (xy - 1)² + (x + y)² + 16 - 8(x+y) = 0

=> (xy -1)² + (x + y - 4)² = 0

as square can not be negative

=>  (xy -1)² = 0 & (x + y - 4)² =0

=> xy = 1   &  x + y = 4  

x³ + y³ = ( x+ y)³ -3xy(x+y)

=> x³ + y³ = 4³ - 3(1)(4)

=> x³ + y³ = 64 - 12

=> x³ + y³ =  52

Answer 2

Question :

If (x²+1)(y²+1)+16 = 8(x+y),

then find the value of x³ + y³

Answer :

x³ + y³ = 52

Step-by-step explanation :

we will expand it and get

x²y² + x² + y² + 1 + 16                                                = 8 ( x + y )

[ x² + y² + 2xy ] - 2xy + 1 + 16 - 8 ( x+y ) + x²y²         = 0

[ ( x + y )² - 8 ( x + y ) + 16 ]  + [ x²y² - 2xy + 1 ]          = 0

[ ( x + y ) - 4]² + [ xy - 1 ]²                                            = 0

so now we get,

x+y = 4 and xy = 1

x³y³ = ( x + y )² - 3xy ( x + y )

       = 4³ - 3 (1) (4)

       = 64 - 12

       = 52