Q: Solve the equation x^{3} – [x]= 5 , where [x] denotes the integral part of the number x.

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Solution: Q x = [x] + f where 0 ≤ f < 1

given equation becomes x^{3} – (x – f) = 5

i.e. x^{3} – x = 5 – f ⇒ 4 < x^{3} – x ≤ 5

Now, x^{3} – x is negative for x ∈ (-∞, -1) ∪ (0, 1)

So, possible values of x lie in the interval [-1, 0] ∪ [1, ∞)

for –1 ≤ x ≤ 0, we have x^{3} – x < 1 < 4 ;

for x = 1, we have x^{3} – x = 0 < 4

further for x ≥ 2 we have

x^{3} – x = x(x^{2} – 1) ≥ 2(4 – 1) = 6 > 5;

Therefore, 1 < x < 2, ⇒ [x] = 1

Now the original equation can be written as x^{3} – 1 = 5 hence x^{3} = 6,

i.e. x = (6)^{1/3}.