If x+y+z=0 show that (x+y+z) cube -27xyz=0
Answers
Hence proved
Solution:
Given that
x+y+z=0x+y+z=0
x ^ { 3 } + y ^ { 3 } + z ^ { 3 } = 3 x y zx
3
+y
3
+z
3
=3xyz
The above expression can be written as like
x ^ { 3 } + y ^ { 3 } + z ^ { 3 } - 3 x y z = ( x + y + z ) \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - x y - y z - z x \right)x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
Substitute the given value x+y+z=0x+y+z=0 in the above equation
Therefore we will get,
x ^ { 3 } + y ^ { 3 } + z ^ { 3 } - 3 x y z = 0 \left( x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - x y - y z - z x \right)x
3
+y
3
+z
3
−3xyz=0(x
2
+y
2
+z
2
−xy−yz−zx)
If any value is multiplied with zero than total value will be equal to zero
\begin{lgathered}\begin{array} { l } { x ^ { 3 } + y ^ { 3 } + z ^ { 3 } - 3 x y z = 0 } \\\\ { x ^ { 3 } + y ^ { 3 } + z ^ { 3 } = 3 x y z } \end{array}\end{lgathered}
x
3
+y
3
+z
3
−3xyz=0
x
3
+y
3
+z
3
=3xyz
Step-by-step explanation:
PLEASE MARK AS A BRAINLIST YOU
