If (x+y)1/3+(y+z)1/3+(z+x)1/3=0 thenshow that (x+y+z)3=9(x3+y3+z3)
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)
giggi , 7 Years ago
Grade 10
10 grade maths> If (x+y) 1/3 +(y+z) 1/3 +(z+x) 1/3 =0 the...
1 Answers
Arun
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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