Question

Solve (1001) taise to power 1/3 by binomial theorem

Answer 1

$(1001) {}^{1 \div 3} \\ (1 + 1000) {}^{ \frac{1}{3} } \\ = by \: formula \\ (1 + x) {}^{n} = 1 + nx + \frac{n{(n - 1)}}{2} x {}^{2} \\ = 1 + \frac{1}{3} 1000 + \frac{1}{3} ( \frac{ - 2}{3} )(1000) {}^{2} \div 2 \\ = 1 + 333.3 - 323.2 = 10.1$

Answer 2

Answer:

Given, \bold{\sqrt[3]{1003}}

3

1003

We know, according to Binomial theorem,

if Any number in the form of (1 + a)ⁿ , where a << 1

Then, (1 + a)ⁿ ≈ 1 + na

Now, (1003)⅓ = (1000 + 3)⅓

= (1000)⅓[ 1 + 3/1000]⅓

= 10[1 + 0.003]⅓

∵ 0.003 << 1

so, 10(1 + 0.003)⅓ = 10(1 + 0.003 × 1/3) = 10(1 + 0.001)

= 10 × 1.001

= 10.01

Hence, \bold{\sqrt[3]{1003}}

3

1003

= 10.01