Using Binomial theorem, evaluate $(96)^3$.

**OR**

Find the middle terms in the expansion of $\left(3 - \dfrac{x^3}{6}\right)^7$.

Question

Using Binomial theorem, evaluate $(96)^3$.

**OR**

Find the middle terms in the expansion of $\left(3 - \dfrac{x^3}{6}\right)^7$.

Accepted Answer

>

> [!TLDR]
> $(96)^3=884{,}736$, computed by expanding $(100-4)^3$.
>
> Write $96=100-4$ and use the binomial expansion $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ with $a=100,\ b=4$:
> $$(96)^3=(100-4)^3=100^3-3(100)^2(4)+3(100)(4)^2-4^3.$$
> Compute each term:
> $$100^3=1{,}000{,}000,$$
> $$3(100)^2(4)=3(10{,}000)(4)=120{,}000,$$
> $$3(100)(4)^2=3(100)(16)=4{,}800,$$
> $$4^3=64.$$
> So:
> $$(96)^3=1{,}000{,}000-120{,}000+4{,}800-64.$$
> $$=880{,}000+4{,}800-64=884{,}800-64=884{,}736.$$
> …

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Q.Using Binomial theorem, evaluate $(96)^3$ . OR Find the middle terms in the expansion of $\left(3 - \dfrac{x^3}{6}\right)^7$ .

Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

The Pattern You Already Know

The Precise Statement

Where Do Those Coefficients Come From?

The General Term

What If It's Not Just $x$ and $y$ ? …

Q.Using Binomial theorem, evaluate (96)3(96)^3(96)3. OR Find the middle terms in the expansion of (3−x36)7\left(3 - \dfrac{x^3}{6}\right)^7(3−6x3​)7.

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Q.Using Binomial theorem, evaluate $(96)^3$ . OR Find the middle terms in the expansion of $\left(3 - \dfrac{x^3}{6}\right)^7$ .

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