Question

A1/3 + b1/3 + c1/3 = 0, then (a + b + c)6 is equal to:

Answer 1

Answer:

see in image

Step-by-step explanation:

(a+b+c)6=0 hence proved

Answer 2

The value of $(a+b+c)^6=0$

Step-by-step explanation:

We have,

$a^{1/3}+b^{1/3}+c^{1/3}=0$

To find, the value of $(a+b+c)^6=?$

∴ $(a^{1/3}+b^{1/3}+c^{1/3})^3$

Using the formula,

$(x+y+z)^3=x^{3}+y^{3}+z^{3}+(x+y+z)(xy+yz+zx)$

∴ $(a+b+c)^6=[(a+b+c)^3]^2$

$=[a^{3}+b^{3}+c^{3}+(a+b+c)(ab+bc+ca)]^2$   ..... (1)

∴ $(a^{1/3}+b^{1/3}+c^{1/3})^3$

$=(a^{\dfrac{1}{3}})^3+(b^{\dfrac{1}{3}})^3+(b^{\dfrac{1}{3}})^3+(a^{\dfrac{1}{3}}+b^{\dfrac{1}{3}}+z^{\dfrac{1}{3}})(a^{\dfrac{1}{3}}b^{\dfrac{1}{3}}+b^{\dfrac{1}{3}}c^{\dfrac{1}{3}}+c^{\dfrac{1}{3}}a^{\dfrac{1}{3}})$

$=a+b+c+(0})(a^{\dfrac{1}{3}}b^{\dfrac{1}{3}}+b^{\dfrac{1}{3}}c^{\dfrac{1}{3}}+c^{\dfrac{1}{3}}a^{\dfrac{1}{3}})$

∴ a + b + c = 0

The value of $(a+b+c)^6=0$

Hence, the value of $(a+b+c)^6=0$