Question

The value of cube root (9+4root 5) + cube root (9-4root 5) is

Answer 1

Answer:

$\bf\red{Verified Answer}$

Step-by-step explanation:

To make the typing easier,

let A = (9 + 4√5)^(1/3) ---> A³ = 9 + 4√5

let B = (9 - 4√5)^(1/3) ---> B³ = 9 - 4√5

Let X = A + B So, now: X = (9 + 4√5)^(1/3) + (9 - 4√5)^(1/3) (the problem)

Also: A·B = [ (9 + 4√5)^(1/3) ] · [ (9 - 4√5)^(1/3) ]

= [ (9 + 4√5) · (9 - 4√5) ] ^(1/3)

= [ 81 - 16 · 5 ] ^(1/3)

= [ 1 ] ^(1/3)

= 1

Since A·B = 1,

A²·B = A(AB) = A(1) = A = (9 + 4√5)^(1/3)

A·B² = (AB)B = (1)B = B = (9 - 4√5)^(1/3)

Sinc X = A + B,

X³ = (A + B)³ = A³ + 3A²·B + 3A·B² + B³

X³ = (9 + 4√5) + [ 3(9 + 4√5)^(1/3) ] + [ 3(9 - 4√5)^(1/3) ] + ( 9 - 4√5 )

Rearranging:

X³ = 18 + 3[ (9 + 4√5)^(1/3) + (9 - 4√5)^(1/3) ]

X³ = 18 + 3[ A + B ]

But, since A + B = X

X³ = 18 + X

X³ - X - 18 = 0

Factoring:

(X - 3)(X² + 3X + 6) = 0

So: X = 3 or X = [-3 ± i√(15) ] / 2

Since the answer is a pure real number, the answer is 3!

Answer 2

Answer:

Rearranging:

X³ = 18 + 3[ (9 + 4√5)^(1/3) + (9 - 4√5)^(1/3) ]

X³ = 18 + 3[ A + B ]

But, since A + B = X

X³ = 18 + X

X³ - X - 18 = 0

Factoring:

(X - 3)(X² + 3X + 6) = 0

So: X = 3 or X = [-3 ± i√(15) ] / 2

Hence, the answer is a pure real number, the answer is 3!