SOLUTION: If x and y are real, x^4y^5+x^5y^4 = 810 & x^6y^3+x^3y^6 = 945, Find 2x^3+(xy)^3+2y^3 = ?

Algebra ->  Real-numbers -> SOLUTION: If x and y are real, x^4y^5+x^5y^4 = 810 & x^6y^3+x^3y^6 = 945, Find 2x^3+(xy)^3+2y^3 = ?      Log On


   



Question 1020225: If x and y are real, x^4y^5+x^5y^4 = 810 & x^6y^3+x^3y^6 = 945, Find 2x^3+(xy)^3+2y^3 = ?
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E4y%5E5%2Bx%5E5y%5E4+=+810+==>x%5E3y%5E3%28x%5E2y%2Bxy%5E2%29+=+810 after factoring x%5E3y%5E3 and changing the order of the terms.
==> x%5E3y%5E3%283x%5E2y%2B3xy%5E2%29+=+2430 (Equation A)
after multiplying both sides by 3.
Similarly, factor out x%5E3y%5E3 from the 2nd equation x%5E6y%5E3%2Bx%5E3y%5E6+=+945:
x%5E3y%5E3%28x%5E3+%2B+y%5E3%29+=+945 (Equation B)
Add corresponding sides of Eqn. A and Eqn. B, to get
x%5E3y%5E3%28x%5E3+%2B+3x%5E2y%2B3xy%5E2%2B+y%5E3%29+=+3375
==>x%5E3y%5E3%28x%2By%29%5E3+=+3375
==> xy(x+y) = 15, (Equation C)
after taking cube roots of both sides.
Now the first equation x%5E4y%5E5%2Bx%5E5y%5E4+=+810+ is the same as
x%5E4y%5E4%28x%2By%29+=+810 (Equation D)
Dividing Eqn.D by Eqn.C, we get
x%5E3y%5E3+=+810%2F15+=54
Hence, +%28xy%29%5E3=54 (Equation E)
Putting x%5E3y%5E3+=+54 into Eqn.B, we get
54%28x%5E3%2By%5E3%29+=+945 ==> x%5E3%2By%5E3+=+35%2F2, or
2x%5E3%2B2y%5E3+=+35 (Equation F)
Finally, combining Equations E and F, we get
2x%5E3%2B%28xy%29%5E3%2B2y%5E3+=+35%2B54+=+highlight%2889%29.