Answer:
a = -1, a=4
Step-by-step explanation:
( a^3 +4a^2 -3) ^1/3 = a+1
We need to cube each side to get rid of the radical
( a^3 +4a^2 -3) ^1/3 ^3 = (a+1)^3
( a^3 +4a^2 -3) = (a+1)^3
Using the formula (x+y)^3=x^3+3x^2y+3xy^2+y^3
we can expand (a+1)^3 to a^3+3a^2+3a+1
( a^3 +4a^2 -3) =a^3+3a^2+3a+1
Moving all the terms to the left hand side
( a^3 +4a^2 -3) -(a^3+3a^2+3a+1) =a^3+3a^2+3a+1-(a^3+3a^2+3a+1)
Distribute the minus sign
( a^3 +4a^2 -3) -a^3-3a^2-3a-1 =0
Combine like terms
a^2 -3a-4 =0
Factor
(a-4) (a+1) =0
Using the zero product property
a-4 =0 a+1=0
a=4 a=-1
Check since we cubed each side for extraneous solutions
( a^3 +4a^2 -3) ^1/3 = a+1
a=4
( 4^3 +4 *4^2 -3) ^1/3 = 4+1
(64 +64 -3) ^1/3 =5
125 ^1/3 =5
5=5 True
a=-1
( (-1)^3 +4 *(-1)^2 -3) ^1/3 = -1+1
(-1 +4 -3) ^1/3 =0
0 ^1/3 =0
0=0 True
