how do i do these types of problems? and what are they called? 2(x+8)^3/4=8
First divide both sides by 2, then you have (x+8)^3/4 = 4
Raise both sides to the 4/3 power
((x+8)^3/4)^4/3 = 4^4/3
x+8 = 6.3496
x= - 1.650395
Solve for x: 1/2 (x+8)^3 = 8 Multiply both sides by 2: (x+8)^3 = 16 Taking cube roots gives 2 2^(1/3) times the third roots of unity: x+8 = -2 (-2)^(1/3) or x+8 = 2 2^(1/3) or x+8 = 2 (-1)^(2/3) 2^(1/3) Subtract 8 from both sides: x = -8-2 (-2)^(1/3) or x+8 = 2 2^(1/3) or x+8 = 2 (-1)^(2/3) 2^(1/3) Subtract 8 from both sides: x = -8-2 (-2)^(1/3) or x = 2 2^(1/3)-8 or x+8 = 2 (-1)^(2/3) 2^(1/3) Subtract 8 from both sides: Answer: | | x = -8-2 (-2)^(1/3) or x = 2 2^(1/3)-8 or x = 2 (-1)^(2/3) 2^(1/3)-8
+129850 2(x+8)^3/4=8 divide both sides by 2
(x + 8)^(3/4) = 4 raise both sides to the 4th power
(x + 8)^3 = 4^4
(x + 8)^3 = 256 take the cube root of both sides
x + 8 = 4*cuberoot(4) subtract 8 from both sides
x = 4*cuberoot(4) - 8 =
4*[ cuberoot(4) - 2] =
4*[ 2^(2/3) - 2 ]
Solve for x: 1/2 (x+8)^3 = 8 Multiply both sides by 2: (x+8)^3 = 16 Taking cube roots gives 2 2^(1/3) times the third roots of unity: x+8 = -2 (-2)^(1/3) or x+8 = 2 2^(1/3) or x+8 = 2 (-1)^(2/3) 2^(1/3) Subtract 8 from both sides: x = -8-2 (-2)^(1/3) or x+8 = 2 2^(1/3) or x+8 = 2 (-1)^(2/3) 2^(1/3) Subtract 8 from both sides: x = -8-2 (-2)^(1/3) or x = 2 2^(1/3)-8 or x+8 = 2 (-1)^(2/3) 2^(1/3) Subtract 8 from both sides: Answer: | | x = -8-2 (-2)^(1/3) or x = 2 2^(1/3)-8 or x = 2 (-1)^(2/3) 2^(1/3)-8
