Find-the-integer-part-of-the-number-2015-2016-2017-1-3-

Question Number 165831 by HongKing last updated on 09/Feb/22

Find the integer part of the number :

\sqrt[3]{2015 \cdot 2016 \cdot 2017}

Commented by mr W last updated on 09/Feb/22

\sqrt[3]{x (x + 1) (x + 2)}

> \sqrt[3]{x \times x \times x} = x

\sqrt[3]{x (x + 1) (x + 2)} = \sqrt[3]{(x + 1) ({(x + 1)}^{2} - 1)}

< \sqrt[3]{(x + 1) {(x + 1)}^{2}} = x + 1

x < \sqrt[3]{x (x + 1) (x + 2)} < x + 1

\Rightarrow [\sqrt[3]{x (x + 1) (x + 2)}] = x

\Rightarrow [\sqrt[3]{2015 \times 2016 \times 2017} = 2015

Answered by naka3546 last updated on 09/Feb/22

Let x = 2016

\sqrt[3]{2015 \cdot 2016 \cdot 2017}

= \sqrt[3]{(x - 1) \cdot x \cdot (x + 1)}

= \sqrt[3]{x (x^{2} - 1)}

= \sqrt[3]{x^{3} - x} < \sqrt[3]{x^{3}}

R e m e m b e r t h a t

x (x + 1) - 3 x + 1 < x (x + 1)

{(x - 1)}^{2} < x (x + 1)

\sqrt[3]{(x - 1) \cdot {(x - 1)}^{2}} < \sqrt[3]{(x - 1) \cdot x \cdot (x + 1)}

2015 = \sqrt[3]{{(x - 1)}^{3}} < \sqrt[3]{x \cdot (x^{2} - 1)} < \sqrt[3]{x^{3}} = 2016

Integer part of \sqrt[3]{2015 \cdot 2016 \cdot 2017} is 2015 .

Commented by naka3546 last updated on 09/Feb/22

y e s, s i r .