Answer:
Step-by-step explanation:
[tex]\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}}\\\\[/tex]
Take cube,
[tex][(x+4)^{\frac{1}{3}}]^{3} =- [(2x +8)^{\frac{1}{3}}^{3}]\\\\(x+4)^{3*\frac{1}{3}}=-(2x + 8)^{3*\frac{1}{3}}\\\\[/tex]
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
