Question

Question in attachment.

Answer 1

Let P = {2a + √(4a²-6²)}⅓
P³ = {2a + √(4a -6²) }
1/P = 1/{2a + √(4a²-6²)}⅓
rationalize
= {2a - √(4a² - 6²)}⅓/{2a-√(4a2-6²)}⅓{2a+√(4a2-6²)}⅓
= {2a - √(4a²-6²)}⅓/[(2a)²-√(4a²-6²)²]⅓
= {2a - √(4a²-6²)}⅓/(4a²-4a²+6²)⅓
= {2a-√(4a²-6²)}⅓/(6²)⅓

so, {2a - √(4a²-6²)}⅓ = 6⅔/P
and {2a - √(4a²-6²)} = 6²/P³
and x = P + 6⅔/P
now, take cube both sides,

x³ = P³ + 6²/P³ + 3.P×6⅔/P( P+6⅔/P)
x³ = {2a + √(4a²-6²)} + {2a -√(4a²-6²)} + 3×6⅔.x
x³ = 4a + 3.6⅔x
x³ - 3.6⅔x - 4a = 0
x³ - 3.6⅔x -4a + 13a = 13a
x³ - 3.6⅔x +9a = 13a