Free help with homework

Why join Brainly?

  • ask questions about your assignment
  • get answers with explanations
  • find similar questions



This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Hi there! I am not so sure if this is the right answer but I will show it:

First we factor out xd from the equation and get:
xd(y+ x^{3} +x y^{2} -y)=0 \\ xd(x^3+xy^2)=0 \\  x^{2} d(x^2+y^2)=0

So our 1st case is that x^2d=0
Case 1.1 x^2=0 \\ x=0
Case 1.2 d=0

Then our second case would be  x^{2} +y^2=0

Remember that all square numbers are greater than or equal to zero therefore y^2 cannot be negative and cannot be positive because that would make  x^{2} negative so y^2=0 therefore:
x^2=-y^2 \\ x=y=0

So our solution set:

0 0 0
The Brain
  • The Brain
  • Helper
Not sure about the answer?
Learn more with Brainly!
Having trouble with your homework?
Get free help!
  • 80% of questions are answered in under 10 minutes
  • Answers come with explanations, so that you can learn
  • Answer quality is ensured by our experts