Verified answer

Let x = (54+30√3)^(1/3) + (54-30√3)^(1/3).

Cube both sides, noting that

(54+30√3)^(1/3) (54-30√3)^(1/3) = 216^(1/3) = 6:

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x^3 = (54+30√3)^(3/3) + 3(54+30√3)^(2/3) (54-30√3)^(1/3)

+ 3(54+30√3)^(1/3) (54-30√3)^(2/3) + (54-30√3)^(3/3)

....= (54+30√3) + 3 * 6 * (54+30√3)^(1/3) + 3 * 6 * (54-30√3)^(1/3) + (54-30√3)

.....= 108 + 18 [(54+30√3)^(1/3) + (54-30√3)^(1/3)]

.....= 108 + 18x

So, x^3 - 18x - 108 = 0.

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Now, we solve for x.

By the Rational Root Theorem (and the polynomial having leading coefficient 1), the only possible rational roots are factors of -108, the constant term.

With trial and error, we find that x = 6 is a root.

So, (x - 6) is a factor of x^3 - 18x - 108 by the Factor Theorem.

Dividing yields x^3 - 18x - 108 = (x - 6)(x^2 + 6x + 18)

So, (x - 6)(x^2 + 6x + 18) = 0

==> x = 6 or -3 ± 3i.

Since x = (54+30√3)^(1/3) + (54-30√3)^(1/3) is real,

we conclude that (54+30√3)^(1/3) + (54-30√3)^(1/3) = 6.

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Double check:

http://www.wolframalpha.com/input/?i=+%2854+%2B30%...

I hope this helps!

(54 +30√(3))^(1/3) + (54 - (30√(3))^(1/3)

= (3 + √3) + (3 - √3)

= 6