If $$ a^{1 / 3}+b^{1 / 3}+c^{1 / 3}=0 $$ sen prove $$ (a+b+c)^{3}=27 a b c $$

Question

UpStudy AI · Accepted Answer

To prove that $$(a + b + c)^3 = 27abc$$ given that $$a^{1/3} + b^{1/3} + c^{1/3} = 0$$, let's proceed step by step.

1. **Define Variables:**
   Let’s set:
   $$
   x = a^{1/3}, \quad y = b^{1/3}, \quad z = c^{1/3}
   $$
   The given condition becomes:
   $$
   x + y + z = 0
   $$

2. **Use the Identity for Sum of Cubes:**
   Recall the algebraic identity:
   $$
   x^3 + y^3 + z^3 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz
   $$
   Since $$x + y + z = 0$$, this simplifies to:
   $$
   x^3 + y^3 + z^3 = 3xyz
   $$

3. **Relate Back to $$a$$, $$b$$, and $$c$$:**
   Substitute back $$x^3 = a$$, $$y^3 = b$$, and $$z^3 = c$$ into the equation:
   $$
   a + b + c = 3xyz = 3(a^{1/3} b^{1/3} c^{1/3}) = 3(abc)^{1/3}
   $$

4. **Cube Both Sides:**
   To find $$(a + b + c)^3$$, cube both sides of the equation:
   $$
   (a + b + c)^3 = \left(3(abc)^{1/3}\right)^3 = 27abc
   $$

Thus, we have shown that:
$$
(a + b + c)^3 = 27abc
$$

**Conclusion:**
Given that $$a^{1/3} + b^{1/3} + c^{1/3} = 0$$, it indeed follows that $$(a + b + c)^3 = 27abc$$.

**Final Answer:**
After expanding, (a + b + c)³ equals 27 abc when a¹⁄³ + b¹⁄³ + c¹⁄³ = 0. Thus,

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

If \( a^{1 / 3}+b^{1 / 3}+c^{1 / 3}=0 \) sen prove \( (a+b+c)^{3}=27 a b c \)

Solución de tutoría real

Responder

Solución

Mind Expander

preguntas relacionadas

Latest Algebra Questions