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Question: Obtain the approximate value of \(\sqrt[3]{{28}}\)....

Obtain the approximate value of 283\sqrt[3]{{28}}.

Explanation

Solution

Cube root can be expressed as the number which produces a given number when cubed. In other words, the cube-root of a number is the factor that is multiplied by it three times to get that number. It is expressed as n3\sqrt[3]{n} and is read as the cube root of the number “n”. For example 273=3×3×33=3\sqrt[3]{{27}} = \sqrt[3]{{3 \times 3 \times 3}} = 3 since here, the number “33 “ is repeated three times. Here we will find the cube root by using the different identities.

Complete step by step solution:
Given expression: 283\sqrt[3]{{28}}
Here the nearest number to the given number whose cube root is possible is 2727
Let us assume that, x=27,δx=1,dx=1x = 27,\delta x = 1,dx = 1
Given expression can be re-written in the form as-
y=x13y = {x^{\dfrac{1}{3}}} …..(A)
Now the above expression can be written as –
y+δy=(x+δx)13(x)13+dxy + \delta y = {(x + \delta x)^{\dfrac{1}{3}}} \approx {(x)^{\dfrac{1}{3}}} + dx …. (B)
y+δy(x)13+dyy + \delta y \approx {(x)^{\dfrac{1}{3}}} + dy
So, we can state that y=(x)13y = {(x)^{\dfrac{1}{3}}} and δydy\delta y \approx dy
Take the differential on both the sides of the equations in the equation (A)
d(y)=d(x)13d(y) = d{(x)^{\dfrac{1}{3}}}
Take differentiation with respect to “x”
dy=13x23dxdy = \dfrac{1}{3}{x^{ - \dfrac{2}{3}}}dx ….. (C)
Place value of equation (C) in the equation (B)
(x+δx)13(x)13+dy=x13+13x23dx{(x + \delta x)^{\dfrac{1}{3}}} \approx {(x)^{\dfrac{1}{3}}} + dy = {x^{\dfrac{1}{3}}} + \dfrac{1}{3}{x^{ - \dfrac{2}{3}}}dx
Place given numbers in the equation (x+δx)13x13+13x23dx{(x + \delta x)^{\dfrac{1}{3}}} \approx {x^{\dfrac{1}{3}}} + \dfrac{1}{3}{x^{ - \dfrac{2}{3}}}dx
(27+1)13(27)13+13(27)23(1){(27 + 1)^{\dfrac{1}{3}}} \approx {(27)^{\dfrac{1}{3}}} + \dfrac{1}{3}{(27)^{ - \dfrac{2}{3}}}(1)
Simplify adding the terms and using the laws of power and exponent.
(28)133+13(33)23{(28)^{\dfrac{1}{3}}} \approx 3 + \dfrac{1}{3}{({3^3})^{ - \dfrac{2}{3}}}
Common factors from the numerator and the denominator cancels each other.
(28)133+13(3)2{(28)^{\dfrac{1}{3}}} \approx 3 + \dfrac{1}{3}{(3)^{ - 2}}
Apply the law of negative exponent law-
(28)133+13(19){(28)^{\dfrac{1}{3}}} \approx 3 + \dfrac{1}{3}\left( {\dfrac{1}{9}} \right)
(28)133+127{(28)^{\dfrac{1}{3}}} \approx 3 + \dfrac{1}{{27}}
Take LCM and simplify –
(28)138227{(28)^{\dfrac{1}{3}}} \approx \dfrac{{82}}{{27}}
Simplify the above equation finding the division –
(28)133.037{(28)^{\dfrac{1}{3}}} \approx 3.037

Note:
We should know the difference between the concepts of cubes and cube-roots and apply it accordingly. When the same natural number is multiplied thrice, we get the cube of that number. It is symbolized as the exponent with the power For example – the cube of “n” number is stated as n3=n×n×n{n^3} = n \times n \times n.