The square root of square of n is 8 less than cube root of n. Find the value of n?
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Square of n = n^2
Square root of Square of n = √n^2
= n
According to the question, Square root of the Square of n = Cube root of n - 8
Thus, n = ∛n - 8
n = 1/n^3 - 8
n = (1 - 8 n^3) / n^3
n^4 = 1 - 8 n^3
n^4 + 8 n^3 = 1
n^3 (n + 8) = 1
n^3 = 1 or n + 8 = 1
n= 1 or n = -7
Thus n can either be 1 or -7
Square root of Square of n = √n^2
= n
According to the question, Square root of the Square of n = Cube root of n - 8
Thus, n = ∛n - 8
n = 1/n^3 - 8
n = (1 - 8 n^3) / n^3
n^4 = 1 - 8 n^3
n^4 + 8 n^3 = 1
n^3 (n + 8) = 1
n^3 = 1 or n + 8 = 1
n= 1 or n = -7
Thus n can either be 1 or -7
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