Maths · Quadratic equations in real and complex number systems and their solutions
ff and evaluate: ^{2}}{x y}+\frac{(2 y+3 z)^{2}}{y z}+ \) ^{2}}{z x} \)
ff \( x+2 y+3 z=0 \) and \( x^{3}+4 y^{3}+ \) \( \mathbf{9} z^{3}=18 x y z ; \) evaluate: \( \frac{(x+2 y)^{2}}{x y}+\frac{(2 y+3 z)^{2}}{y z}+ \) \( \frac{(3 z+x)^{2}}{z x} \)
- A. 18
- B. 23
- C. 16
- D. 11
Step-by-step solution
Given x+2y+3z=0, we have x+2y = -3z, 2y+3z = -x, 3z+x = -2y. Substituting into the expression yields (9z^2)/(xy) + x^2/(yz) + 4y^2/(zx) = (x^3+4y^3+9z^3)/(xyz). Using the second condition x^3+4y^3+9z^3 = 18xyz, the expression simplifies to 18.
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