Respuesta :

Answer:

[tex]\sqrt[3]{62}[/tex] is closest to 4.

Step-by-step explanation:

Our approach consist in comparing the given cubic root with respect to other two cubic roots consecutive to each other and whose result is an integer. The following condition must be satisfied:

[tex]n < \sqrt[3]{62} < n+1[/tex]

[tex]n^{3} < 62 < (n+1)^{3}[/tex]

If [tex]n = 3[/tex], then the simultaneous inequation is observed:

[tex]3^{3}<62< 4^{3}[/tex]

[tex]27 < 62 < 64[/tex]

Since 62 is closer to the upper bound, then [tex]\sqrt[3]{62}[/tex] is closest to 4.

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