Question

Obtain the approximate value of $\sqrt[3]{328}$.

Answer 1

The nearest number to 28 whose cube root is positive integer is 27.

Let $x=27,\delta x=1,dx=1$

$y=x^{\frac{1}{3}}$ (1)

$y+\delta y=(x+\delta x{)}^{\frac{1}{3}}\approx (x{)}^{\frac{1}{3}}+dx$ (2)

$y+\delta y\approx (x{)}^{\frac{1}{3}}+dy$

So $y=x^{\frac{1}{3}}$ and $\delta y\approx dy$

Now taking differential on both sides of (1)

$d\left(y\right)=d\left(x^{\frac{1}{3}}\right)$

$dy=\frac{1}{3}{x}^{\frac{-2}{3}}dx$ (3)

Substituting (3) in (2)

$(x+\delta x{)}^{\frac{1}{3}}\approx (x{)}^{\frac{1}{3}}+dy=(x{)}^{\frac{1}{3}}+\frac{1}{3}{x}^{\frac{-2}{3}}dx$

Substituting numerical values in $(x+\delta x{)}^{\frac{1}{3}}\approx (x{)}^{\frac{1}{3}}+\frac{1}{3}{x}^{\frac{-2}{3}}dx$

$(27+1{)}^{\frac{1}{3}}\approx (27{)}^{\frac{1}{3}}+\frac{1}{3}(27{)}^{\frac{-2}{3}}\times 1$

$(28{)}^{\frac{1}{3}}\approx 3+\frac{1}{3}(3^3{)}^{\frac{-2}{3}}$

$(28{)}^{\frac{1}{3}}\approx 3+\frac{1}{3}\times \frac{1}{9}$

$(28{)}^{\frac{1}{3}}\approx 3+\frac{1}{27}=\frac{82}{27}=3.037$

Hence $(28{)}^{\frac{1}{3}}\approx 3.037$