∛{7+5√2} + ∛{7-5√2} =?
namku:
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let ∛(7 +5√2) = a = x ( 1 + y√2)
and ∛(7 - 5√2) = b
So 7 + 5√2 = x³ (1 + y √2)³
= x³ (1 + 6 y²) + √2 * x³ (3 y+ 2 y³)
equate rational and irrational parts.
(1 + 6 y²) = 7 / x³ --- (1)
(3 y + 2 y³) = 5 / x³ ---- (2)
Solving these two equations we get:
14 y³ - 30 y² + 21 y - 5 = 0
looking at the coefficients : 14+21-30-5=0
so y = 1 is a solution. then we get: 14 y² - 16 y + 5 as the other factor. The roots of this polynomial are imaginary.
Thus we get y = 1. hence by (1), 1 + 6 *1² =7/x³ => x = 1
so ∛(7 + 5√2) = 1 + √2
similarly ∛(7 - 5√2) = 1 - √2
the answer is 2.
and ∛(7 - 5√2) = b
So 7 + 5√2 = x³ (1 + y √2)³
= x³ (1 + 6 y²) + √2 * x³ (3 y+ 2 y³)
equate rational and irrational parts.
(1 + 6 y²) = 7 / x³ --- (1)
(3 y + 2 y³) = 5 / x³ ---- (2)
Solving these two equations we get:
14 y³ - 30 y² + 21 y - 5 = 0
looking at the coefficients : 14+21-30-5=0
so y = 1 is a solution. then we get: 14 y² - 16 y + 5 as the other factor. The roots of this polynomial are imaginary.
Thus we get y = 1. hence by (1), 1 + 6 *1² =7/x³ => x = 1
so ∛(7 + 5√2) = 1 + √2
similarly ∛(7 - 5√2) = 1 - √2
the answer is 2.
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