how did u type the cube root ? can uh elp me even i want to typw but i dont know

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1.99995

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how did u type the cube root ? can uh elp me even i want to typw but i dont know

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1.99995

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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.