Question

if (y+z)^1/3 (z+x)^1/3 +(x+y)^1/3 =0 show that (x+y+z)^3 =9(x^3 +y^3+z^3)

Answer 1

Answer:

Step-by-step explanation:

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x⅓ + y⅓ + z⅓ = 0

Let x⅓ = a

y⅓ = b

z⅓ = c

then , a + b + c = 0

we know, when, a + b + c = 0

then, a³ + b³ + c³ = 3abc

hence,

(x⅓)³ + (y⅓)³ + (z⅓)³ = 3(x⅓)(y⅓)(z⅓)

x + y + z = 3(xyz)⅓

take cube both sides,

(x + y + z)³ = (3)³{ (xyz)⅓}³

(x + y + z)³ = 27xyz(answer )