Question

the value of (997)⅓ according to binomial theoram

Answer 1

The answer cannot come according to binomial theorem. Because binomial theorem holds integral value of n. But 1/3 is not an integer.

Answer 2

Answer: $(997)^{\frac{1}{3}} = 9.99$

Explanation:

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

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

Using binomial theorem

$(1-x)^n=1-nx+\dfrac{n(n-1)}{2}x^2-\dfrac{n(n-1)(n-2)}{6}x^3......$

$(1-\dfrac{3}{1000})^{\frac{1}{3}} = 1-\dfrac{1}{3}\times\dfrac{3}{1000}+\dfrac{\dfrac{1}{3}(\dfrac{1}{3}-1)}{2}(\dfrac{3}{1000})^{2}.........$

$(1-0.003)^{\frac{1}{3}}= 0.999$

$10(1-\dfrac{3}{1000})^{\frac{1}{3}} = 10\times 0.999$

$10(1-\dfrac{3}{1000})^{\frac{1}{3}} =9.99$

$(997)^{\frac{1}{3}} = 9.99$

Hence, This is the required solution