Solve for "a" in this radical equation.

45 points!!

Please show all work

a = -1 and a = 4

Step-by-step explanation:

∛a³ + 4a² -3 = a + 1    

a³ + 4a² - 3 = (a + 1)³

a³ + 4a² -3 = a³ + 3a² + 3a + 1

a² - 3a -4

(a + 1) (a - 4)

a + 1 = 0

a = -1

a -4 = 0

a = 4

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

step-by-step explanation:

b and c