SOLUTION: Solve. (k + 4)2/3 + 6(k+4)1/3 + 8=0

Question 728875: Solve.
(k + 4)2/3 + 6(k+4)1/3 + 8=0
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Solve.
(k + 4)2/3 + 6(k+4)1/3 + 8=0
let u=(k+4)^(1/3)
u^2=(k+4)^(2/3)
..
u^2+6u+8=0
(u+2)(u+4)=0
..
u=-2
-2=(k+4)^(1/3)
cube both sides
-8=k+4
k=-12
..
u=-4
-4=(k+4)^(1/3)
cube both sides
-64=k+4
k=-68
check:
for k=-12
(k + 4)2/3 + 6(k+4)1/3 + 8
(-8)2/3 + 6(-8)1/3 + 8
4+(-12)+8
4-12+8=0
..
for k=-68
(k + 4)2/3 + 6(k+4)1/3 + 8
(-64)2/3 + 6(-64)1/3 + 8
16+(-24)+8
16-24+8=0
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