$x^{3} = 729$ . Find x

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Solution

Basically you need to find the cube root of 729.
Prime factorization of 729 yields = 3^6
So, cube root of $729 = 729^{\frac{1}{3}} = {3^{6}}^{\frac{1}{3}} = 3^{2} = 9$
So, x = 9

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If x+y+z=9 and xy+yz+zx = 23, the value of $(x^{3} + y^{3} + z^{3} - 3 x y z) = ?$

(a) 108

(b) 207

(d) 729

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