a) Find $(a+b)^4 - (a-b)^4$. (2)
b) Hence evaluate $(\sqrt{3}+\sqrt{2})^4 - (\sqrt{3}-\sqrt{2})^4$. (1)

Question

Accepted Answer

>

> [!TLDR]
> Expand both fourth powers by the binomial theorem, subtract — the even-power terms cancel and only the odd-power terms survive, giving a compact factored form. Then substitute $a=\sqrt3,\,b=\sqrt2$.
>
> **a) Expanding**
>
> $$(a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
> $$(a-b)^4 = a^4-4a^3b+6a^2b^2-4ab^3+b^4$$
>
> Subtracting, the $a^4$, $6a^2b^2$, and $b^4$ terms cancel:
>
> $$(a+b)^4-(a-b)^4 = 8a^3b+8ab^3 = 8ab(a^2+b^2)$$
> …

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.a) Find $(a+b)^4 - (a-b)^4$ .

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.a) Find (a+b)4−(a−b)4(a+b)^4 - (a-b)^4(a+b)4−(a−b)4.

🎓Unlock everything free for 14 days

Q.a) Find $(a+b)^4 - (a-b)^4$ .

Unlock everything free for 14 days