Q.(a) Prove that $\sqrt[3]{x^3+6}-\sqrt[3]{x^3+3}$ is approximately equal to $\dfrac{1}{x^2}$ when $x$ is sufficiently large. OR

Theorem 5.1's binomial theorem is stated only for a positive integer exponent $n$. But the pattern extends — as an infinite series, not a finite expansion — to ANY rational (indeed any real) exponent, provided $|x|<1$.

Theorem 5.4 (Binomial series for rational exponent). For any rational number $n$,

(Unlike Theorem 5.1, this is an infinite series, and it is stated in this course without proof.) Replacing $x\to-x$ or $n\to-n$ produces the companion forms

(all valid for $|x|<1$). Four special cases are worth memorising outright: $(1+x)^{-1}=1-x+x^2-\cdots$, $(1-x)^{-1}=1+x+x^2+\cdots$, $(1-x)^{-2}=1+2x+3x^2+4x^3+\cdots$, $(1+x)^{-2}=1-2x+3x^2-4x^3+\cdots$.

Working technique for a general binomial $(A+Bx)^n$. Factor out $A^n$ to reduce to the standard form: $(A+Bx)^n=A^n\left(1+\tfrac BA x\right)^n$, expand $\left(1+\tfrac BA x\right)^n$ by the theorem, and the validity condition becomes $\left|\tfrac BA x\right|<1$, i.e. $|x|<\left|\tfrac AB\right|$.

Numerical root approximations. To approximate $\sqrt[q]{N}$ for $N$ close to a perfect $q^{th}$ power $M^q$: write $N=M^q(1+h)$ with $h=\tfrac{N-M^q}{M^q}$ small, so $\sqrt[q]N=M(1+h)^{1/q}\approx M\left(1+\tfrac hq\right)$ — keeping just the first two (or three) terms of the series gives a fast, accurate decimal approximation (e.g. $\sqrt[3]{65}\approx4.02$, $\sqrt[3]{1001}\approx10.00$). …