2.1.5Algebra — Introduction & Intermediate
Algebraic identities — (a+b)², (a−b)², (a+b)(a−b), (a+b)³, (a−b)³, (a³+b³), (a³−b³)
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The Seven Core Identities
Identity 1:
Derivation from First Principles:
(a+b)^2 &= (a+b)(a+b) && \text{Definition of squaring} \\ &= a(a+b) + b(a+b) && \text{Distributive property on first factor} \\ &= a \cdot a + a \cdot b + b \cdot a + b \cdot b && \text{Distributive property on each term} \\ &= a^2 + ab + ab + b^2 && \text{Simplify multiplication} \\ &= a^2 + 2ab + b^2 && \text{Combine like terms} \end{align}$$ **WHY this works**: Each term in the first binomial multiplies each term in the second. The middle term $2ab$ appears because $ab$ comes from both $a \times b$ and $b \times a$. **Geometric Interpretation**: ![[2.1.05-Algebraic-identities-—-(a+b)2,-(a−b)2,-(a+b)(a−b),-(a+b)3,-(a−b)3,-(a3+b3),-(a3−b3).png]] The square with side $(a+b)$ has area $(a+b)^2$, which equals one $a^2$ square, one $b^2$ square, and two $ab$ rectangles. >[!example] Worked Example 1: $(x+3)^2$ >**Identify**: Here $a = x$, $b = 3$ >**Apply formula**: >$$\begin{align} >(x+3)^2 &= x^2 + 2(x)(3) + 3^2 \\ >&= x^2 + 6x + 9 >\end{align}$$ >**Why this step?** We substituted $a=x$ and $b=3$ into the identity, then simplified $2 \cdot x \cdot 3 = 6x$ and $3^2 = 9$. > >**Verify by direct multiplication**: >$(x+3)(x+3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9$ ✓ >[!example] Worked Example 2: $(2y+5)^2$ >**Identify**: $a = 2y$, $b = 5$ >$$\begin{align} >(2y+5)^2 &= (2y)^2 + 2(2y)(5) + 5^2 \\ >&= 4y^2 + 20y + 25 >\end{align}$$ > >**Why this step?** $(2y)^2 = 2^2 \cdot y^2 = 4y^2$ (power of a product), and $2\cdot 2y \cdot 5 = 20y$. --- ### Identity 2: $(a-b)^2 = a^2 - 2ab + b^2$ >[!definition] Square of a Difference >When you ==square a binomial difference==, you get the square of the first term, minus twice the product of both terms, plus the square of the second term. **Derivation**: $$\begin{align} (a-b)^2 &= (a-b)(a-b) \\ &= a(a-b) - b(a-b) && \text{Distributive property} \\ &= a^2 - ab - ba + b^2 && \text{Expand each product} \\ &= a^2 - ab + b^2 && \text{Since } ba = ab \\ &= a^2 - 2ab + b^2 && \text{Combine like terms} \end{align}$$ **Key difference from Identity 1**: The middle term is $-2ab$ (negative) because we're subtracting $b$. >[!example] Worked Example 3: $(x-4)^2$ >$a = x$, $b = 4$ > >$$\begin{align} >(x-4)^2 &= x^2 - 2(x)(4) + 4^2 \\ >&= x^2 - 8x + 16 >\end{align}$$ > >**Common mistake check**: Notice the last term is $+16$, not $-16$. The $b^2$ is always positive. >[!mistake] Common Error: $(a-b)^2 = a^2 - b^2$ >**Why it feels right**: Students think "square distributes over subtraction" or forget the middle term. >**Steel-man**: This seems logical if you treat the square like multiplication distributes, but squaring is **not** a linear operation. $(a-b)^2$ means $(a-b)$ times itself, which produces a middle term. > >**The fix**: Remember the identity has **three terms**, and visualize the geometric square or use FOIL as a check. The middle term $-2ab$ is unavoidable. > >**Counter-example**: Let $a=5$, $b=3$. Then $(5-3)^2 = 2^2 = 4$, but $5^2 - 3^2 = 25-9=16 \neq 4$. --- ### Identity 3: $(a+b)(a-b) = a^2 - b^2$ >[!definition] Difference of Squares >The product of a sum and difference of the same terms equals the ==difference of their squares==. **Derivation**: $$\begin{align} (a+b)(a-b) &= a(a-b) + b(a-b) && \text{Distributive property} \\ &= a^2 - ab + ba - b^2 && \text{Expand each product} \\ &= a^2 - ab + ab - b^2 && \text{Since } ba = ab \\ &= a^2 - b^2 && \text{Middle terms cancel!} \end{align}$$ **WHY this is powerful**: The middle terms $-ab$ and $+ab$ cancel perfectly, leaving a clean result. This identity is ==reversible== and used heavily in factoring. >[!example] Worked Example 4: $(x+7)(x-7)$ >$a = x$, $b = 7$ > >$(x+7)(x-7) = x^2 - 7^2 = x^2 - 49$ > >**No middle term!** That's the beauty of this identity. >[!example] Worked Example 5: Factoring $9y^2 - 16$ >**Recognize pattern**: $9y^2 = (3y)^2$ and $16 = 4^2$ > >$$9y^2 - 16 = (3y)^2 - 4^2 = (3y+4)(3y-4)$$ > >**Why this step?** We identified $a=3y$ and $b=4$, then applied the identity in reverse (factoring instead of expanding). --- ### Identity 4: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ >[!definition] Cube of a Sum >When you ==cube a binomial sum==, you get four terms with coefficients 1, 3, 3, 1 (the same as row 3 of Pascal's Triangle). **Derivation**: $$\begin{align} (a+b)^3 &= (a+b)(a+b)^2 && \text{Definition: } (a+b)^3 = (a+b) \cdot (a+b)^2 \\ &= (a+b)(a^2 + 2ab + b^2) && \text{Use Identity 1} \\ &= a(a^2 + 2ab + b^2) + b(a^2 + 2ab + b^2) && \text{Distributive property} \\ &= a^3 + 2a^2b + ab^2 + ba^2 + 2ab^2 + b^3 && \text{Multiply each term} \\ &= a^3 + 2a^2b + ab^2 + 2ab^2 + b^3 && \text{Rearrange } ba^2 = a^2b \\ &= a^3 + 3a^2b + 3ab^2 + b^3 && \text{Combine like terms} \end{align}$$ > **Pattern**: Powers of a decrease from 3 to 0, powers of b increase from 0 to 3, and coefficients are 1, 3, 1. > > [!example] Worked Example 6: (x+2)^3 > a = x, b = 2 > $$\begin{align} >(x+2)^3 &= x^3 + 3x^2(2) + 3x(2)^2 + 2^3 \\ >&= x^3 + 6x^2 + 3x(4) + 8 \\ >&= x^3 + 6x^2 + 12x + 8 >\end{align}$$ >**Why each step?** We substitute values, then simplify: $3x^2 \cdot 2 = 6x^2$, $3x \cdot 4 = 12x$, $2^3 = 8$. >[!mnemonic] Remember the coefficients: "1-3-3-1" >Think of a **pyramid**: one block on top, three blocks in the next row, three blocks below that, one at the bottom. Or remember **Pascal's Triangle** row 3. --- ### Identity 5: $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ >[!definition] Cube of a Difference >Similar to the sum, but with ==alternating signs==: positive, negative, positive, negative. **Derivation**: $$\begin{align} (a-b)^3 &= (a-b)(a-b)^2 \\ &= (a-b)(a^2 - 2ab + b^2) && \text{Use Identity 2} \\ &= a(a^2 - 2ab + b^2) - b(a^2 - 2ab + b^2) \\ &= a^3 - 2a^2b + ab^2 - ba^2 + 2ab^2 - b^3 \\ &= a^3 - 2a^2b - a^2b + ab^2 + 2ab^2 - b^3 \\ &= a^3 - 3a^2b + 3ab^2 - b^3 \end{align}$$ > **Sign pattern**: + − + −, matching the alternating structure of subtraction. > > [!example] Worked Example 7: (2y-1)^3 > a = 2y, b = 1 > $$\begin{align} >(2y-1)^3 &= (2y)^3 - 3(2y)^2(1) + 3(2y)(1)^2 - 1^3 \\ >&= 8y^3 - 3(4y^2) + 3(2y) - 1 \\ >&= 8y^3 - 12y^2 + 6y - 1 >\end{align}$$ > >**Why this step?** $(2y)^3 = 8y^3$ (cube each factor), $3 \cdot 4y^2 = 12y^2$, $3 \cdot 2y = 6y$. --- ### Identity 6: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ >[!definition] Sum of Cubes >The ==sum of two cubes== factors into a binomial sum times a trinomial. **Derivation from RHS**: Let's expand $(a+b)(a^2 - ab + b^2)$ and verify it equals $a^3 + b^3$: $$\begin{align} (a+b)(a^2 - ab + b^2) &= a(a^2 - ab + b^2) + b(a^2 - ab + b^2) \\ &= a^3 - a^2b + ab^2 + ba^2 - ab^2 + b^3 \\ &= a^3 - a^2b + ab^2 - ab^2 + b^3 \\ &= a^3 + b^3 \end{align}$$ > **Key insight**: The middle terms cancel when you expand, leaving only the cubes. > > **WHY the trinomial has that form**: The trinomial $a^2 - ab + b^2$ is designed so that when multiplied by $(a+b)$, the $\pm a^2b$ and $\pm ab^2$ terms cancel. > > [!example] Worked Example 8: Factor x^3 + 27 > **Recognize**: 27 = 3^3, so this is x^3 + 3^3 > a = x, b = 3 > $$\begin{align} >x^3 + 27 &= (x+3)(x^2 - x \cdot 3 + 3^2) \\ >&= (x+3)(x^2 - 3x + 9) >\end{align}$$ > **Verify**: Multiply back to check: > (x+3)(x^2-3x+9) = x^3 - 3x^2 + 9x + 3x^2 - 9x + 27 = x^3 + 27 ✓ > > [!example] Worked Example 9: Factor 8y^3 + 125 > **Recognize**: 8y^3 = (2y)and 125 = 5^3 > a = 2y, b = 5 > $$8y^3 + 125 = (2y+5)((2y)^2 - (2y)(5) + 5^2) = (2y+5)(4y^2 - 10y + 25)$$ > a^3 - b^3 = (a-b)(a^2 + ab + b^2) > $$\begin{align} (a-b)(a^2 + ab + b^2) &= a(a^2 + ab + b^2) - b(a^2 + ab + b^2) \\ &= a^3 + a^2b + ab^2 - ba^2 - ab^2 - b^3 \\ &= a^3 + a^2b - a^2b + ab^2 - ab^2 - b^3 \\ &= a^3 - b^3 \end{align}$$ > **Contrast with Identity 6**: The trinomial has +ab (not -ab) because we're factoring a difference, not a sum. > > [!example] Worked Example 10: Factor x^3 - 64 > 64 = 4^3, so a = x, b = 4 > $$x^3 - 64 = (x-4)(x^2 + x \cdot 4 + 4^2) = (x-4)(x^2 + 4x + 16)$$ >[!mistake] Common Error: Confusing Signs in Sum/Difference of Cubes >**The mistake**: Writing $a^3 + b^3 = (a+b)(a^2 + ab + b^2)$ or $a^3 - b^3 = (a-b)(a^2 - ab + b^2)$ > >**Why it feels right**: Students think the signs should "match" between the binomial and trinomial. > >**Steel-man**: It's natural to expect symmetry. But the identities are designed so that middle terms cancel during expansion. The **opposite sign** in the middle of the trinomial is necessary for cancellation. > >**The fix**: Remember "**SOAP**" for sum/difference of cubes: >- **S**ame sign (binomial sign matches the original problem) >- **O**posite sign (middle term of trinomial is opposite) >- **A**lways **P**ositive (last term of trinomial is always $+b^2$) >For $a^3 + b^3$: $(a+b)(a^2 - ab + b^2)$ — Same +, Opposite −, Always Positive >For $a^3 - b^3$: $(a-b)(a^2 + ab + b^2)$ — Same −, Opposite +, Always Positive --- ## Summary Table | Identity | Formula | Key Feature | |---|---| | $(a+b)^2$ | $a^2 + 2ab + b^2$ | Middle term is $+2ab$ | | $(a-b)^2$ | $a^2 - 2ab + b^2$ | Middle term is $-2ab$ | | $(a+b)(a-b)$ | $a^2 - b^2$ | Middle terms cancel | | $(a+b)^3$ | $a^3 + 3a^2b + 3ab^2 + b^3$ | Coefficients: 1-3-3-1 | | $(a-b)^3$ | $a^3 - 3a^2b + 3ab^2 - b^3$ | Alternating signs | | $a^3 + b^3$ | $(a+b)(a^2 - ab + b^2)$ | SOAP: Same, Opposite, Always + | | $a^3 - b^3$ | $(a-b)(a^2 + ab + b^2)$ | SOAP: Same, Opposite, Always + | --- ## Applications >[!example] Worked Example 11: Simplify $(x+5)^2 - (x-5)^2$ >**Method 1 — Expand both**: >$$\begin{align} >(x+5)^2 - (x-5)^2 &= (x^2 + 10x + 25) - (x^2 - 10x + 25) \\ >&= x^2 + 10x + 25 - x^2 + 10x - 25 \\ >&= 20x >\end{align}$$ > **Method 2 — Use difference of squares**: > Let A = (x+5) and B = (x-5). Then A^2 - B^2 = (A+B)(A-B): > $$\begin{align} >(x+5)^2 - (x-5)^2 &= [(x+5) + (x-5)][(x+5) - (x-5)] \\ >&= [x+5+x-5][x+5-x+5] \\ >&= (2x)(10) \\ >&= 20x >\end{align}$$ > >**Why Method 2 is better**: Fewer steps, less room for sign errors. >[!example] Worked Example 12: Evaluate $103^3 - 3(103)^2(3) + 3(103)(3)^2 - 3^3$ >**Recognize the pattern**: This matches $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a=103$, $b=3$ > >$$103^3 - 3(103)^2(3) + 3(103)(3)^2 - 3^3 = (103-3)^3 = 100^3 = 1,\!000,\!000$$ > >**Why this step?** Instead of computing huge products, we recognized the identity and collapsed it to $(100)^3$. --- >[!recall]- Feynman Technique: Explain to a 12-Year-Old >Imagine you have a square piece of paper with side length $(a+b)$. What's its area? > >Instead of just saying "$(a+b)$ times $(a+b)$," let's cut it up! Draw a line $a$ units from the left edge, and another line $a$ units from the bottom. Now you have four rectangles: >- A small square in the corner, $a \times a = a^2$ >- A tall skiny rectangle on the right, $a \times b = ab$ >- A wide flat rectangle on top, $b \times a = ab$ >- A tiny square in the top-right, $b \times b = b^2$ > >Add them up: $a^2 + ab + ab + b^2 = a^2 + 2ab + b^2$. That's why $(a+b)^2$ isn't just $a^2 + b^2$ — you're missing those two rectangles in the middle! > >For $(a+b)(a-b)$, imagine you have a rectangle that's $a$ units wide. If you add $b$ units to the length, you get area $a(a+b)$. But if you subtract $b$ units, you get $a(a-b)$. When you multiply $(a+b)$ and $(a-b)$, the middle parts cancel out like magic, leaving just $a^2 - b^2$. --- ## Connections - [[Distributive Property]] — the foundation of all these identities - [[FOIL Method]] — a technique for expanding binomials - [[Factoring Polynomials]] — these identities are used in reverse - [[Pascal's Triangle]] — explains the coefficients in $(a+b)^n$ - [[Binomial Theorem]] — the general case for any power - [[Completing the Square]] — uses $(a+b)^2$ to solve quadratics - [[Difference of Squares in Number Theory]] — used in factoring and divisibility profs - [[Quadratic Equations]] — discriminant patterns emerge from $(a-b)^2$ --- ## #flashcards/maths What is the expanded form of $(a+b)^2$? :: $a^2 + 2ab + b^2$ What is the expanded form of $(a-b)^2$? ::: $a^2 - 2ab + b^2$ What is $(a+b)(a-b)$ equal to? ::: $a^2 - b^2$ (difference of squares) What is the middle term in $(a+b)^2$? ::: $2ab$ (positive, coefficient 2) What is the middle term in $(a-b)^2$? ::: $-2ab$ (negative, coefficient 2) Expand $(a+b)^3$ ::: $a^3 + 3a^2b + 3ab^2 + b^3$ Expand $(a-b)^3$ ::: $a^3 - 3a^2b + 3ab^2 - b^3$ Factor $a^3 + b^3$ ::: $(a+b)(a^2 - ab + b^2)$ Factor $a^3 - b^3$ ::: $(a-b)(a^2 + ab + b^2)$ What are the coefficients in $(a+b)^3$? ::: 1, 3, 3, 1 (Pascal's Triangle row 3) What is the SOAP mnemonic for sum of cubes? ::: Same sign in binomial, Opposite sign in trinomial middle term, Always Positive last term In $(a+b)(a-b) = a^2 - b^2$, why do the middle terms vanish? ::: Because $+ab$ and $-ab$ cancel when you expand using the distributive property True or False: $(a-b)^2 = a^2 - b^2$ :: False. $(a-b)^2 = a^2 - 2ab + b^2$ (the middle term $-2ab$ is essential) Factor $x^2 - 49$ ::: $(x+7)(x-7)$ using $a^2 - b^2 = (a+b)(a-b)$ Factor $27x^3 + 8$ ::: $(3x+2)(9x^2 - 6x + 4)$ using sum of cubes, where $27x^3 = (3x)^3$ and $8 = 2^3$ What is $(x+5)^2 - (x-5)^2$? :: $20x$ (use difference of squares or expand both) In the trinomial factor of $a^3 - b^3$, what is the sign of the middle term? ::: Positive: $a^2 + ab + b^2$ ## 🖼️ Concept Map ```mermaid flowchart TD DIST[Distributive property] SUMSQ["(a+b)² = a²+2ab+b²"] DIFSQ["(a−b)² = a²−2ab+b²"] DIFSQR["(a+b)(a−b) = a²−b²"] SUMCUBE["(a+b)³ = a³+3a²b+3ab²+b³"] DIFCUBE["(a−b)³ = a³−3a²b+3ab²−b³"] SUMCUBES["a³+b³ = (a+b)(a²−ab+b²)"] DIFCUBES["a³−b³ = (a−b)(a²+ab+b²)"] GEO[Geometric area model] SHORT[Expansion shortcuts] DIST -->|derives| SUMSQ DIST -->|derives| DIFSQ DIST -->|derives| DIFSQR SUMSQ -->|sign flip on b| DIFSQ SUMSQ -->|geometric proof| GEO SUMSQ -->|extends to cube| SUMCUBE DIFSQ -->|extends to cube| DIFCUBE SUMCUBE -->|sign flip| DIFCUBE SUMCUBES -->|factors| SUMCUBE DIFCUBES -->|factors| DIFCUBE SUMSQ -->|serve as| SHORT DIFSQR -->|serve as| SHORT ``` ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > ![[audio/2.1.05-Algebraic-identities-—-(a+b)2,-(a−b)2,-(a+b)(a−b),-(a+b)3,-(a−b)3,-(a3+b3),-(a3−b3).mp3]] ## 🔊 Hinglish (regional understanding) > [!intuition]- Hinglish mein samjho > Dekho beta, ye algebraic identities basically shortcuts hain jo tumhara time bachate hain. Jab tumhe baar-baar $(a+b)(a+b)$ jaise expressions ko multiply karna padta hai, tab har baar poora multiplication karne ki jagah tum seedha formula laga sakte ho — jaise $(a+b)^2 = a^2 + 2ab + b^2$. Ye koi jaadu nahi hai, ye toh simple distributive property se hi nikalte hain. Middle term $2ab$ isliye aata hai kyunki $a \times b$ aur $b \times a$ dono baar milta hai, samajh gaye? Bilkul waise jaise square ka area side ka square hota hai — tum ek-ek unit ginte nahi ho, direct formula lagate ho. > > Ab ye important kyun hai? Kyunki maths mein aage jaake — factorization, quadratic equations, simplification — har jagah ye identities kaam aayenge. Agar tumhe ye zabaani yaad aur samajh mein aa gaye, toh tumhari speed aur accuracy dono badh jaayegi. Sabse common galti jo students karte hain woh hai $(a-b)^2 = a^2 - b^2$ samajhna — ye bilkul galat hai! Yaad rakho, $(a-b)^2$ mein teen term hote hain: $a^2 - 2ab + b^2$. Ek chhota sa check karo: $a=5, b=3$ lo, toh $(5-3)^2 = 4$ hota hai, lekin $25 - 9 = 16$ — dono barabar nahi, matlab galat formula. > > Mera advice ye hai ki inhe sirf ratto mat, balki geometric picture ya FOIL method se ek baar khud derive karke dekho. Jab tumhe pata hoga ki formula aaya kahan se, tab exam mein tension mein bhi galti nahi karoge. Ye saat identities tumhare algebra ki foundation hain, toh inhe achhe se pakad lo — baaki chapters mein ye tumhare best dost ban jaayenge.