Question

the value of 997^1/3 according to binomial theorem is

Answer 1

Answer: The value of $(997)^{\frac{1}{3}}$ is $10.0099$

Explanation:

Given that,

$(997)^{\dfrac{1}{3}}=(1000-3)^{\frac{1}{3}}$

$(1000-3)^{\dfrac{1}{3}}= (10(1-0.003)^{\frac{1}{3}})$

$(10(1-0.003)^{\dfrac{1}{3}})= 1+\dfrac{1}{3}\times0.003+\dfrac{\frac{1}{3}(\frac{1}{3}-1)}{2}\times(0.003)^{2}......$

$(10(1-0.003)^{\frac{1}{3}})= 10.0099$

$(997)^{\frac{1}{3}}= 10.0099$

Hence, The value of $(997)^{\frac{1}{3}}$ is $10.0099$.

Answer 2

Answer:

the value of (997)$^{1/3}$ is 10.0099

Explanation:

$(997)^{1/3}$  = $(1000-3)^{1/3}$

$(1000-3)^{1/3}$ = { 10 (1-0.003)$^{1/3}$ }

{10(1-0.003)$^{1/3}$ } =  1+1/3×0.003 + $\frac{1}{3} (\frac{1}{3} - 1)/2$ ×(0.003)²....

{10(1-0.003)$^{1/3}$ } =10.0099

( 997)$^{1/3}$ =10.0099

the value of (997)$^{1/3}$ is 10.0099

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