Grade 11Algebra

Find x^1/3 + y^1/3 + z^1/3 = 0
Then find (x + y + z)^3

aditya kukade

12 Years agoGrade 11

3 Answers

Arun Kumar

12 Years ago

Hello Student

$\\$Let $x^{1/3} = a, y^{1/3} = b, z^{1/3} = c \\$The question then becomes, if $(a+b+c) = 0 \\$ then what is the value of $(a^3+b^3+c^3)^3. \\$We have a+b $= -c. \\=> (a+b)^3 = (-c)^3 = -c^3. \\=> a^3+b^3+3*a*(b^2)+3*(a^2)*b = -c^3 \\=> a^3+b^3+c^3 = -{ 3*a*(b^2)+3*(a^2)*b } = -3*a*b*(a+b) = -3*a*b*(-c) = 3abc. \\=> (a^3+b^3+c^3)^3 = (3abc)^3 = 27(a^3)(b^3)(c^3) = 27xyz.$

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty

Pallavi Das

12 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

Rishi Sharma

5 Years ago

Dear Student,
Please find below the solution to your problem.

we know if a + b + c = 0 then
a^3 + b^3 + c^3 = 3abc
so if x^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

Thanks and Regards