Question

What is the solution to the equation?

Answer 1

Answer:

Step-by-step explanation:

$\sqrt[3]{x+4}+\sqrt[3]{2x +8}=0\\\\\sqrt[3]{x+4}= -\sqrt[3]{2x + 8}\\\\(x +4)^{\frac{1}{3}}=-(2x+8)^{\frac{1}{3}}$

Take cube,

$[(x+4)^{\frac{1}{3}}]^{3} =- [(2x +8)^{\frac{1}{3}}^{3}]\\\\(x+4)^{3*\frac{1}{3}}=-(2x + 8)^{3*\frac{1}{3}}$

x + 4 = - (2x + 8)

x +4 = 2x *(-1) + 8 *(-1)

x+ 4 = -2x - 8

x +2x+4 = - 8

3x + 4 = - 8

      3x = - 8 -4

      3x = -12

        x = -12/3

x = -4