What is px^{2}+qy^{2}+rz^{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}.
- A. (x^{2}+y^{2}+z^{2})(p+q+r) ✓
- B. -(x^{2}+y^{2}+z^{2})(p+q+r)
- C. (y^{2}+z^{2}-x^{2})(r-q-p)
- D. (x^{2}+y^{2}-z^{2})(p-q-r)
Correct Answer: A. (x^{2}+y^{2}+z^{2})(p+q+r)
Explanation
Substituting p, q, r, the expression px^2+qy^2+rz^2 expands to x^6+y^6+z^6-3x^2y^2z^2. Using the standard algebraic identity for sum of cubes A^3+B^3+C^3-3ABC, this factors into (x^2+y^2+z^2)(x^4+y^4+z^4-x^2y^2-y^2z^2-z^2x^2), which is precisely (x^2+y^2+z^2)(p+q+r).
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