If \(x-y+z=0\), find the value of \(\displaystyle\frac{y^2}{2xz}-\frac{x^2}{2yz}-\frac{z^2}{2xy}\).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{3}{2}\)
If three numbers add up to zero, the sum of their cubes equals three times their product: \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\). Here the three numbers are \(x\), \(-y\) and \(z\).
Taking the common denominator \(2xyz\): \(\displaystyle\frac{y^2}{2xz}-\frac{x^2}{2yz}-\frac{z^2}{2xy}=\frac{y^3-x^3-z^3}{2xyz}\)
\(x+(-y)+z=0\), so \(x^3+(-y)^3+z^3=3\cdot x\cdot(-y)\cdot z\), that is \(x^3-y^3+z^3=-3xyz\)
Multiplying by \(-1\): \(y^3-x^3-z^3=3xyz\)
Value \(\displaystyle=\frac{3xyz}{2xyz}=\frac{3}{2}\)