(7x-2)^(1/3) + (7x+5)^(1/3) = 3

∛(7x-2) + ∛(7x+5) = 3

Let a = ∛(7x-2) b = ∛(7x+5)

We thus have b³ = (7x+5) = 7x -2 +7 = a³ + 7

So a + b = 3

b = 3 - a ,then

b³ = (3 - a)³ = a³ + 7

3³ - a³ -3(3)(-a)(3 - a) = a³ + 7

27 - a³ +9a(3 - a) = a³ + 727 - a³ + 27a - 9a² = a³ + 72a³ - 9a² + 27a -20 = 02a³ - 2a² - 7a² + 7a + 20a - 20 = 02a²(a - 1) - 7a(a - 1) +20(a - 1) = 0(a - 1)(2a² - 7a +20) = 0(a - 1)(a² - 7/2a +10) = 0(a - 1)(a² – 2×a×(7/4) + (7/4)² – 49/16 + 160/16) = 0(a - 1)[(a – 7/4)² + 111/16] = 0Thus the only real solution is a = 1then a = ∛(7x - 2) = 1 a³ = (7x - 2) = 1 7x = 3 HENCE x = 3/7

27 - a³ +9a(3 - a) = a³ + 7

27 - a³ + 27a - 9a² = a³ + 7

2a³ - 9a² + 27a -20 = 0

2a³ - 2a² - 7a² + 7a + 20a - 20 = 0

2a²(a - 1) - 7a(a - 1) +20(a - 1) = 0

(a - 1)(2a² - 7a +20) = 0

(a - 1)(a² - 7/2a +10) = 0

(a - 1)(a² – 2×a×(7/4) + (7/4)² – 49/16 + 160/16) = 0

(a - 1)[(a – 7/4)² + 111/16] = 0

Thus the only real solution is a = 1

then a = ∛(7x - 2) = 1

a³ = (7x - 2) = 1

7x = 3

HENCE x = 3/7

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