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We know that: $(x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)…...(i)$

Given that: $x=\frac{5}{12},y=\frac{-3}{4},z=\frac{1}{3}$

  • $(x+y)=\frac{5-9}{12}=-1/3$
  • $(y+z)=\frac{-9+4}{12}=-5/12$
  • $(z+x)=\frac{4+5}{12}=3/4$

Put all these value in equation (I), we get:

$\implies0=x^3+y^3+z^3+3*(-1/3)*(-5/12)*3/4$

$\implies x^3+y^3+z^3=\frac{-5}{16}$

Option $(D)$ is correct.

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