X^2+y^2+z^2=xy+yz+zx then the value of (x^2/yz)+(y^2/zx)+(z^2/xy) is ?
Answers
Answer:
hence proved
Step-by-step explanation:
+y+z=6
xy+yz+zx=12
2) We know that,
\begin{lgathered}{(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + zx) \\ = > {6}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2 \times 12 \\ = > {x}^{2} + { y }^{2} + {z}^{2} = 36 - 24 = 12\end{lgathered}
(x+y+z)
2
=x
2
+y
2
+z
2
+2(xy+yz+zx)
=>6
2
=x
2
+y
2
+z
2
+2×12
=>x
2
+y
2
+z
2
=36−24=12
3) And,
We also know that,
\begin{lgathered}{x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z) \\ \times ( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) \\ = > 6 \times (12 - 12) = 0 \\ = > {x}^{3} + {y}^{3} + {z}^{3} - 3xyz = 0 \\ = > {x}^{3} + {y}^{3} + {z}^{3} = 3xyz\end{lgathered}
x
3
+y
3
+z
3
−3xyz=(x+y+z)
×(x
2
+y
2
+z
2
−xy−yz−zx)
=>6×(12−12)=0
=>x
3
+y
3
+z
3
−3xyz=0
=>x
3
+y
3
+z
3
=3xyz
Hence Proved.