Question

Find the approximate value of $\sqrt[3]{26.96}$

Answer 1

Answer:

$\sqrt[3]{26.96}=2.99852\:(approx.)$

Step-by-step explanation:

$\sqrt[3]{26.96}$

$=(26.96)^\frac{1}{3}$

$=(27-0.04)^\frac{1}{3}$

$=(27)^\frac{1}{3}(1-\frac{0.04}{27})^\frac{1}{3}$

$=(3^3)^\frac{1}{3}(1-\frac{0.04}{27})^\frac{1}{3}$

$=3(1-\frac{0.04}{27})^\frac{1}{3}$

Using, binomial theorem for rational index

$\boxed{(1+x)^n=1+nx+\frac{n(n-1)}{2}+.......}$

$=3[1+\frac{1}{3}(-\frac{0.04}{27})+.............]\:\text{( neglecting the higher powers )}$

$=3-\frac{0.04}{27}$

$=3-\frac{4}{27*100}$

$=3-\frac{0.148}{100}$

$=3-0.00148$

$=2.99852\:(approximately)$

Answer 2

Answer:

Approximate value is 2.99852

Step-by-step explanation:

$\sqrt[3]{26.96}$

= $(26.96)^\frac{1}{3}$

= $(27-0.04)^\frac{1}{3}$

= $(27)^\frac{1}{3}(1-\frac{0.04}{27})^\frac{1}{3}$

= $(3^3)^\frac{1}{3}(1-\frac{0.04}{27})^\frac{1}{3}$

= $3(1-\frac{0.04}{27})^\frac{1}{3}$

Using binomial theorem for rational index

$\boxed{(1+x)^n=1+nx+\frac{n(n-1)}{2}+.......}$

= $3[1+\frac{1}{3}(-\frac{0.04}{27})+.............]\:\text{( neglecting the higher powers )}$

= $3-\frac{0.04}{27}$

= $3-\frac{4}{27*100}$

= $3-\frac{0.148}{100}$

= 3-0.00148

= 2.99852(approximately)

Approximate value is 2.99852