Identity Usage

Using the identity:

$\alpha^3+\beta^3+\gamma^3 = (\alpha+\beta+\gamma)( \alpha^2 +\beta^2 +\gamma^2-\alpha \beta - \beta \gamma - \gamma \alpha)+3\alpha \beta \gamma$

Now substituting $\alpha = a^{\frac{1}{3}} \quad \beta = b^{\frac{1}{3}}\quad \gamma = c^{\frac{1}{3}}$ , we have:

$a+b+c = (a^{\frac{1}{3}}+ b^{\frac{1}{3}} + c^{\frac{1}{3}}) (a^{\frac{2}{3}}+ b^{\frac{2}{3}} + c^{\frac{2}{3}} - a^{\frac{1}{3}} b^{\frac{1}{3}} - b^{\frac{1}{3}} c^{\frac{1}{3}} - c^{\frac{1}{3}}a^{\frac{1}{3}} ) \\ \quad \quad \quad \quad \quad + 3 a^{\frac{1}{3}} b^{\frac{1}{3}} c^{\frac{1}{3}}$

$\quad \quad \quad \quad \quad = 0 + 3(1) = 3$

$\Rightarrow (a+b+c)^3 = 3^3 = \boxed{27}$

$a^{\frac{1}{3}}+b^{\frac{1}{3}}=-c^{\frac{1}{3}}$ ........... $1$

Cubing both sides

$a+b+3a^{\frac{1}{3}}b^{\frac{1}{3}}(a^{\frac{1}{3}}+b^{\frac{1}{3}})=-c$

From $1$

$a+b+3a^{\frac{1}{3}}b^{\frac{1}{3}}(-c^{\frac{1}{3}})=-c$

$a+b+c=3(abc)^{\frac{1}{3}}$

$a+b+c=3$ [as $a\times b\times c=1$ ]

$(a+b+c)^{3}=\boxed{27}$

Put

a = x^3 , b = y^3 , c = z^3 ...........(1)

Then

x + y + z = 0 ...................................... (2)

Since

abc = 1

Then, using Eq,s(1)

(xyz)^3 = 1

Then

xyz = 1 .................................................(3)

From Eq (2)

x + y = -z .............................................(4)

Cubing both sides of Eq (4)

x^3 + 3xy(x + y) + y^3 = -z^3

x^3 + 3xy(-z) + y^3 = -z^3

Then

x^3 + y^3 + z^3 = 3xyz = 3(1) = 3

Then

a + b + c = 3

So

(a + b + c)^3 = 3^3 = 27

27 is the answer. x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-xz) (classical fatoration). Then, if x+y+z=0 --> x³+y³+z³=3xyz. Consider a^1/3=x, b^1/3=y, c^1/3=z. then we have: a^1/3+b^1/3+c^1/3=0 --> a+b+c=3(abc)^1/3=3(1)^1/3=3. Finally, (a+b+c)³=3³=27.