How to show that √(29+6√6)+3√(21-6√6)=10√2?

Let √(29 + 6√6) = √a + √b, where a > b
[√(29 + 6√6)]² = [√a + √b]²
29 + 6√6 = a + b + 2√(ab)
29 + 2√54 = (a + b) + 2√(ab)
(27 + 2) + 2√(27 × 2) = (a + b) + 2√(ab)
Hence, a = 27 and b = 2
√(29 + 6√6) = √27 + √2
√(29 + 6√6) = 3√3 + √2

Let √(21 - 6√6) = √c - √d, where c > d
[√(21 - 6√6)]² = [√c - √d]²
21 - 6√6 = c + d - 2√(cd)
21 - 2√(54) = c + d - 2√(cd)
(18 + 3) - 2√(18 × 3) = c + d - 2√(cd)
Hence, c = 18 and d = 3
√(21 - 6√6) = √18 - √3
√(21 - 6√6) = 3√2 - √3

L.H.S.
= √(29 + 6√6) + 3(21 - 6√6)
= (3√3 + √2) + 3(3√2 - √3)
= 3√3 + √2 + 9√2 - 3√3
= 10√2
= R.H.S.

Hence, √(29 + 6√6) + 3(21 - 6√6) = 10√2