x+y+z=0,3y²+x²+z²/2y²-xz=?
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Step-by-step explanation:
Givenx+y+z=0−−−(1)
\begin{gathered} LHS=\frac{x^{2}}{yz}+\frac{y^{2}}{zx}+\frac{z^{2}}{xy}\\=\frac{x^{3}+y^{3}+z^{3}}{xyz}\\=\frac{(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)+3xyz}{xyz}\\=\frac{0\times (x^{2}+y^{2}+z^{2}-xy-yz-zx)+3xyz}{xyz}\end{gathered}
LHS=
yz
x
2
+
zx
y
2
+
xy
z
2
=
xyz
x
3
+y
3
+z
3
=
xyz
(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)+3xyz
=
xyz
0×(x
2
+y
2
+z
2
−xy−yz−zx)+3xyz
/* From (1) */
=\frac{3xyz}{xyz}=
xyz
3xyz
\begin{gathered}= 3\\=RHS\end{gathered}
=3
=RHS
Therefore.,
\begin{gathered} If\: x+y+z=0 \: then \\\frac{x^{2}}{yz}+\frac{y^{2}}{zx}+\frac{z^{2}}{xy}=3\end{gathered}
Ifx+y+z=0then
yz
x
2
+
zx
y
2
+
xy
z
2
=3
•••♪
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