What is x^{2}(px^{2}+qy^{2}+rz^{2})+qr-p^{2} equal to?

Consider the following for the next two (02) items that follow : Let p=x^{4}-y^{2}z^{2}, q=y^{4}-z^{2}x^{2}, r=z^{4}-x^{2}y^{2}.

  1. A. 0
  2. B. 1
  3. C. p+q+r
  4. D. x^{2}+y^{2}+z^{2}

Correct Answer: A. 0

Explanation

Calculating qr - p^2 simplifies to -x^2(x^6+y^6+z^6-3x^2y^2z^2). Since px^2+qy^2+rz^2 = x^6+y^6+z^6-3x^2y^2z^2, multiplying it by x^2 gives x^2(x^6+y^6+z^6-3x^2y^2z^2). Adding these two results in exactly 0.

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