If x - y + z = 0, then find the value of $$\frac{{{y^2}}}{{2xz}} - \frac{{{x^2}}}{{2yz}} - \frac{{{z^2}}}{{2xy}}?$$
A. $$\frac{3}{2}$$
B. $$\frac{1}{2}$$
C. $$-6$$
D. $$ - \frac{3}{2}$$
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \frac{{{y^2}}}{{2xz}} - \frac{{{x^2}}}{{2yz}} - \frac{{{z^2}}}{{2xy}} \cr & \frac{{{y^3} - {x^3} - {z^3}}}{{2xyz}}\,......\,\left( 1 \right) \cr & x - y + z = 0 \cr & x + z = y \cr & {\text{Cubing both side}} \cr & {\left( {x + z} \right)^3} = {y^3} \cr & {x^3} + {z^3} + 3\left( {x + z} \right)\left( x \right)\left( z \right) = {y^3} \cr & {x^3} + {z^3} + 3\left( y \right)\left( x \right)\left( z \right) = {y^3} \cr & 3xyz = {y^3} - {x^3} - {z^3} \cr & {\text{Put in equation }}\left( 1 \right) \cr & \frac{{3xyz}}{{2xyz}} = \boxed{\frac{3}{2}} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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