Question

I WILL DONATE MY 40 POINTS IF YOU ANSWER RHIS QUESTION..if x^1/3+y^1/3=z^1/3 then (x+y-z)³+27xyz. ​

Answer 1

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

I hope this will help you

Answer 2

Answer:

x1 / 3 + y1 / 3 = z1 / 3 ...(i)

Cubing both sides,

( x1 / 3 + y1 / 3 = z1 / 3 )3 = z

⇒ x + y + 3 x1 / 3 . y1 / 3 ( 1 / 3 + y1 / 3 ) = z

[∴ (a + b)3 = a3 + b3 + 3ab (a + b)]

⇒ x + y – z = - 3 x1 / 3 . y1 / 3 .z1 / 3 ...(ii) [From equation (i)]

∴ (x + y – z)3 + 27xyz = [ - 3x1 / 3 . y1 / 3 . z1 / 3 ]3 + 27xyz [From equation (ii)]

= – 27xyz + 27xyz = 0

Step-by-step explanation:

Hope this helps you.