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

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Accepted Answer

> [!TLDR]
> The key idea is to expand both $(a+b)^4$ and $(a-b)^4$ using the Binomial Theorem, subtract them, and notice that all even-powered terms cancel, leaving only odd-powered terms. The simplified expression is $8a^3b + 8ab^3$. Substituting $a = \sqrt{3}$ and $b = \sqrt{2}$ gives the final value $40\sqrt{6}$.

The Binomial Theorem tells us how to expand any power of a binomial sum. For $(a+b)^4$, the expansion is:
$$(a+b)^4 = \binom{4}{0}a^4 + \binom{4}{1}a^3b + \binom{4}{2}a^2b^2 + \binom{4}{3}ab^3 + \binom{4}{4}b^4$$
which simplifies to:
$$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$

Similarly, for $(a-b)^4$, we replace $b$ with $-b$:
$$(a-b)^4 = a^4 + 4a^3(-b) + 6a^2(-b)^2 + 4a(-b)^3 + (-b)^4$$
$$= a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$$

Now, when we subtract $(a-b)^4$ from $(a+b)^4$, the terms with even powers of $b$ (which are $a^4$, $6a^2b^2$, and $b^4$) appear in both expansions with the same sign, so they cancel out. The terms with odd powers of $b$ (the $a^3b$ and $ab^3$ terms) have opposite signs, so they add up.

> [!WARNING]
> A common mistake is to forget that $(-b)^4 = b^4$ (positive) but $(-b)^3 = -b^3$ (negative). Always check the sign of each term carefully when expanding $(a-b)^n$.

Let's do the subtraction step by step:

1. **Write the expansions side by side:**
   $$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$$
   $$(a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$$

2. **Subtract term by term:**
   $$(a+b)^4 - (a-b)^4 = (a^4 - a^4) + (4a^3b - (-4a^3b)) + (6a^2b^2 - 6a^2b^2) + (4ab^3 - (-4ab^3)) + (b^4 - b^4)$$

3. **Simplify each pair:**
   - $a^4 - a^4 = 0$
   - $4a^3b + 4a^3b = 8a^3b$
   - $6a^2b^2 - 6a^2b^2 = 0$
   - $4ab^3 + 4ab^3 = 8ab^3$
   - $b^4 - b^4 = 0$

4. **Result:**
   $$(a+b)^4 - (a-b)^4 = 8a^3b + 8ab^3$$ …

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

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.Find (a+b)4−(a−b)4(a + b)^4 - (a - b)^4(a+b)4−(a−b)4. Hence, evaluate (3+2)4−(3−2)4\left(\sqrt{3} + \sqrt{2}\right)^4 - \left(\sqrt{3} - \sqrt{2}\right)^4(3​+2​)4−(3​−2​)4.

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Q.Find $(a + b)^4 - (a - b)^4$ . Hence, evaluate $\left(\sqrt{3} + \sqrt{2}\right)^4 - \left(\sqrt{3} - \sqrt{2}\right)^4$ .

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