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If (x+y) 1/3 +(y+z) 1/3 +(z+x) 1/3 =0 then show that (x+y+z) 3 =9(x 3 +y 3 +z 3 )

If (x+y)1/3+(y+z)1/3+(z+x)1/3=0 then
show that (x+y+z)3=9(x3+y3+z3)

Grade:10

1 Answers

Arun
25763 Points
2 years ago
 
we know if a+b+c=0 then a^3+b^3+c^3=3abc
so ifx^1/3 + y^1/3 + z^1/3 = 0 then
x+y+z=3.x^1/3.y^1/3.z^1/3
so (x+y+z)^3=(3.x^1/3.y^1/3.z^1/3)^3
(x+y+z)^3=27xyz
 

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