Algebraic identities — (a+b)², (a−b)², (a+b)(a−b), (a+b)³, (a−b)³, (a³+b³), (a³−b³)

2.1.5 · Maths › Algebra — Introduction & Intermediate

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The Seven Core Identities

Identity 1: $(a+b{)}^2=a^2+2ab+b^2$

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}$$ **YE KYU KAAM KARTA HAI**: Pehle binomial ka har term doosre ke har term se multiply hota hai. Middle term $2ab$ isliye aata hai kyunki $ab$ dono jagah se aata hai — $a \times b$ se bhi aur $b \times a$ se bhi. **Geometric Interpretation**: ![[2.1.05-Algebraic-identities-—-(a+b)2,-(a−b)2,-(a+b)(a−b),-(a+b)3,-(a−b)3,-(a3+b3),-(a3−b3).png]] $(a+b)$ side wale square ka area $(a+b)^2$ hota hai, jo ek $a^2$ square, ek $b^2$ square, aur do $ab$ rectangles ke barabar hota hai. >[!example] Worked Example 1: $(x+3)^2$ >**Identify**: Yahan $a = x$, $b = 3$ >**Apply formula**: >$$\begin{align} >(x+3)^2 &= x^2 + 2(x)(3) + 3^2 \\ >&= x^2 + 6x + 9 >\end{align}$$ >**Ye step kyun?** Humne $a=x$ aur $b=3$ ko identity mein substitute kiya, phir $2 \cdot x \cdot 3 = 6x$ aur $3^2 = 9$ simplify kiya. > >**Direct multiplication se verify karo**: >$(x+3)(x+3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9$ ✓ >[!example] Worked Example 2: $(2y+5)^2$ >**Identify**: $a = 2y$, $b = 5$ >$$\begin{align} >(2y+5)^2 &= (2y)^2 + 2(2y)(5) + 5^2 \\ >&= 4y^2 + 20y + 25 >\end{align}$$ > >**Ye step kyun?** $(2y)^2 = 2^2 \cdot y^2 = 4y^2$ (power of a product), aur $2\cdot 2y \cdot 5 = 20y$. --- ### Identity 2: $(a-b)^2 = a^2 - 2ab + b^2$ >[!definition] Square of a Difference >Jab tum ==ek binomial difference ko square karte ho==, to tumhe milta hai pehle term ka square, minus dono terms ke product ka do guna, plus doosre term ka square. **Derivation**: $$\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}$$ **Identity 1 se key difference**: Middle term $-2ab$ hai (negative) kyunki hum $b$ subtract kar rahe hain. >[!example] Worked Example 3: $(x-4)^2$ >$a = x$, $b = 4$ > >$$\begin{align} >(x-4)^2 &= x^2 - 2(x)(4) + 4^2 \\ >&= x^2 - 8x + 16 >\end{align}$$ > >**Common mistake check**: Dhyan do ki aakhri term $+16$ hai, $-16$ nahi. $b^2$ hamesha positive hota hai. >[!mistake] Common Error: $(a-b)^2 = a^2 - b^2$ >**Ye sahi kyun lagta hai**: Students sochte hain "square subtraction par distribute hota hai" ya middle term bhool jaate hain. >**Steel-man**: Ye logical lagta hai agar tum square ko aise treat karo jaise multiplication distribute hoti hai, lekin squaring ek **linear operation nahi** hai. $(a-b)^2$ ka matlab hai $(a-b)$ khud se multiply, jo ek middle term produce karta hai. > >**The fix**: Yaad rakho ki identity mein **teen terms** hote hain, aur geometric square visualize karo ya check ke liye FOIL use karo. Middle term $-2ab$ inevitable hai. > >**Counter-example**: Maano $a=5$, $b=3$. To $(5-3)^2 = 2^2 = 4$, lekin $5^2 - 3^2 = 25-9=16 \neq 4$. --- ### Identity 3: $(a+b)(a-b) = a^2 - b^2$ >[!definition] Difference of Squares >Same terms ke sum aur difference ka product unke ==squares ke difference== ke barabar hota hai. **Derivation**: $$\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}$$ **YE POWERFUL KYU HAI**: Middle terms $-ab$ aur $+ab$ perfectly cancel ho jaate hain, ek clean result dete hue. Ye identity ==reversible== hai aur factoring mein bahut use hoti hai. >[!example] Worked Example 4: $(x+7)(x-7)$ >$a = x$, $b = 7$ > >$(x+7)(x-7) = x^2 - 7^2 = x^2 - 49$ > >**Koi middle term nahi!** Yahi is identity ki khoobsoorti hai. >[!example] Worked Example 5: $9y^2 - 16$ ko factor karo >**Pattern pehchano**: $9y^2 = (3y)^2$ aur $16 = 4^2$ > >$$9y^2 - 16 = (3y)^2 - 4^2 = (3y+4)(3y-4)$$ > >**Ye step kyun?** Humne $a=3y$ aur $b=4$ identify kiya, phir identity ko reverse mein apply kiya (expand karne ki jagah factor kiya). --- ### Identity 4: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ >[!definition] Cube of a Sum >Jab tum ==ek binomial sum ko cube karte ho==, to tumhe 1, 3, 3, 1 coefficients wale chaar terms milte hain (Pascal's Triangle ki row 3 jaisa). **Derivation**: $$\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}$$ > **Pattern**: $a$ ki powers 3 se 0 tak ghatti hain, $b$ ki powers 0 se 3 tak badhti hain, aur coefficients 1, 3, 1 hain. > > [!example] Worked Example 6: (x+2)^3 > a = x, b = 2 > $$\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}$$ >**Har step kyun?** Hum values substitute karte hain, phir simplify karte hain: $3x^2 \cdot 2 = 6x^2$, $3x \cdot 4 = 12x$, $2^3 = 8$. >[!mnemonic] Coefficients yaad rakho: "1-3-3-1" >Ek **pyramid** socho: upar ek block, usse neeche teen blocks, usse neeche teen blocks, aur neeche ek. Ya **Pascal's Triangle** ki row 3 yaad rakho. --- ### Identity 5: $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ >[!definition] Cube of a Difference >Sum ke jaisa hi, lekin ==alternating signs== ke saath: positive, negative, positive, negative. **Derivation**: $$\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}$$ > **Sign pattern**: + − + −, subtraction ki alternating structure se match karta hai. > > [!example] Worked Example 7: (2y-1)^3 > a = 2y, b = 1 > $$\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}$$ > >**Ye step kyun?** $(2y)^3 = 8y^3$ (har factor ko cube karo), $3 \cdot 4y^2 = 12y^2$, $3 \cdot 2y = 6y$. --- ### Identity 6: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ >[!definition] Sum of Cubes >==Do cubes ka sum== ek binomial sum aur ek trinomial ke product mein factor hota hai. **Derivation from RHS**: $(a+b)(a^2 - ab + b^2)$ ko expand karke verify karte hain ki ye $a^3 + b^3$ ke barabar hai: $$\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}$$ > **Key insight**: Expand karne par middle terms cancel ho jaate hain, sirf cubes bachte hain. > > **Trinomial ki woh form kyun hai**: Trinomial $a^2 - ab + b^2$ is tarah design kiya gaya hai ki jab $(a+b)$ se multiply ho, to $\pm a^2b$ aur $\pm ab^2$ terms cancel ho jayein. > > [!example] Worked Example 8: x^3 + 27 ko factor karo > **Pehchano**: 27 = 3^3, to ye x^3 + 3^3 hai > a = x, b = 3 > $$\begin{align} >x^3 + 27 &= (x+3)(x^2 - x \cdot 3 + 3^2) \\ >&= (x+3)(x^2 - 3x + 9) >\end{align}$$ > **Verify karo**: Wapas multiply karke check karo: > (x+3)(x^2-3x+9) = x^3 - 3x^2 + 9x + 3x^2 - 9x + 27 = x^3 + 27 ✓ > > [!example] Worked Example 9: 8y^3 + 125 ko factor karo > **Pehchano**: 8y^3 = (2y)^3 aur 125 = 5^3 > a = 2y, b = 5 > $$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}$$ > **Identity 6 se contrast**: Trinomial mein +ab hai (−ab nahi) kyunki hum difference factor kar rahe hain, sum nahi. > > [!example] Worked Example 10: x^3 - 64 ko factor karo > 64 = 4^3, to a = x, b = 4 > $$x^3 - 64 = (x-4)(x^2 + x \cdot 4 + 4^2) = (x-4)(x^2 + 4x + 16)$$ >[!mistake] Common Error: Sum/Difference of Cubes mein Signs Confuse Karna >**Galti**: $a^3 + b^3 = (a+b)(a^2 + ab + b^2)$ ya $a^3 - b^3 = (a-b)(a^2 - ab + b^2)$ likhna > >**Ye sahi kyun lagta hai**: Students sochte hain ki signs binomial aur trinomial mein "match" hone chahiye. > >**Steel-man**: Symmetry expect karna natural hai. Lekin identities is tarah design ki gayi hain ki expand karne par middle terms cancel ho jayein. Trinomial ke middle mein **opposite sign** cancellation ke liye zaroori hai. > >**The fix**: Sum/difference of cubes ke liye "**SOAP**" yaad rakho: >- **S**ame sign (binomial sign original problem se match karta hai) >- **O**pposite sign (trinomial ka middle term opposite hai) >- **A**lways **P**ositive (trinomial ka last term hamesha $+b^2$ hota hai) >$a^3 + b^3$ ke liye: $(a+b)(a^2 - ab + b^2)$ — Same +, Opposite −, Always Positive >$a^3 - b^3$ ke liye: $(a-b)(a^2 + ab + b^2)$ — Same −, Opposite +, Always Positive --- ## Summary Table | Identity | Formula | Key Feature | |---|---|---| | $(a+b)^2$ | $a^2 + 2ab + b^2$ | Middle term $+2ab$ hai | | $(a-b)^2$ | $a^2 - 2ab + b^2$ | Middle term $-2ab$ hai | | $(a+b)(a-b)$ | $a^2 - b^2$ | Middle terms cancel ho jaate hain | | $(a+b)^3$ | $a^3 + 3a^2b + 3ab^2 + b^3$ | Coefficients: 1-3-3-1 | | $(a-b)^3$ | $a^3 - 3a^2b + 3ab^2 - b^3$ | Alternating signs | | $a^3 + b^3$ | $(a+b)(a^2 - ab + b^2)$ | SOAP: Same, Opposite, Always + | | $a^3 - b^3$ | $(a-b)(a^2 + ab + b^2)$ | SOAP: Same, Opposite, Always + | --- ## Applications >[!example] Worked Example 11: $(x+5)^2 - (x-5)^2$ simplify karo >**Method 1 — Dono expand karo**: >$$\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}$$ > **Method 2 — Difference of squares use karo**: > Maano A = (x+5) aur B = (x-5). To A^2 - B^2 = (A+B)(A-B): > $$\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}$$ > >**Method 2 better kyun hai**: Kam steps, sign errors ke liye kam jagah. >[!example] Worked Example 12: $103^3 - 3(103)^2(3) + 3(103)(3)^2 - 3^3$ evaluate karo >**Pattern pehchano**: Ye $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ se match karta hai jahan $a=103$, $b=3$ > >$$103^3 - 3(103)^2(3) + 3(103)(3)^2 - 3^3 = (103-3)^3 = 100^3 = 1,\!000,\!000$$ > >**Ye step kyun?** Bade products calculate karne ki jagah, humne identity pehchani aur use $(100)^3$ mein collapse kar diya. --- >[!recall]- Feynman Technique: Ek 12-Saal ke Bacche ko Explain Karo >Socho tumhare paas kagaz ka ek square piece hai jiska side $(a+b)$ hai. Uska area kya hai? > >Sirf "$(a+b)$ times $(a+b)$" kehne ki jagah, ise kaat lo! Baayein edge se $a$ units door ek line khiincho, aur neeche se $a$ units door ek aur line. Ab tumhare paas chaar rectangles hain: >- Ek chota square ek corner mein, $a \times a = a^2$ >- Ek tall skinny rectangle daayein, $a \times b = ab$ >- Ek wide flat rectangle upar, $b \times a = ab$ >- Ek tiny square top-right mein, $b \times b = b^2$ > >Sabko jodo: $a^2 + ab + ab + b^2 = a^2 + 2ab + b^2$. Isliye $(a+b)^2$ sirf $a^2 + b^2$ nahi hota — beech ke do rectangles miss ho rahe hain! > >$(a+b)(a-b)$ ke liye, socho tumhare paas $a$ units wide ek rectangle hai. Agar tum length mein $b$ units jodoge, to area $a(a+b)$ milega. Lekin agar $b$ units ghataoge, to $a(a-b)$ milega. Jab tum $(a+b)$ aur $(a-b)$ multiply karte ho, to beech ke hisse jaadu ki tarah cancel ho jaate hain, sirf $a^2 - b^2$ bachta hai. --- ## Connections - [[Distributive Property]] — in sabhi identities ki neev - [[FOIL Method]] — binomials expand karne ki ek technique - [[Factoring Polynomials]] — ye identities reverse mein use hoti hain - [[Pascal's Triangle]] — $(a+b)^n$ mein coefficients explain karta hai - [[Binomial Theorem]] — kisi bhi power ke liye general case - [[Completing the Square]] — quadratics solve karne ke liye $(a+b)^2$ use karta hai - [[Difference of Squares in Number Theory]] — factoring aur divisibility proofs mein use hota hai - [[Quadratic Equations]] — discriminant patterns $(a-b)^2$ se nikalte hain --- ## #flashcards/maths $(a+b)^2$ ki expanded form kya hai? :: $a^2 + 2ab + b^2$ $(a-b)^2$ ki expanded form kya hai? ::: $a^2 - 2ab + b^2$ $(a+b)(a-b)$ kiske barabar hai? ::: $a^2 - b^2$ (difference of squares) $(a+b)^2$ mein middle term kya hai? ::: $2ab$ (positive, coefficient 2) $(a-b)^2$ mein middle term kya hai? ::: $-2ab$ (negative, coefficient 2) $(a+b)^3$ expand karo ::: $a^3 + 3a^2b + 3ab^2 + b^3$ $(a-b)^3$ expand karo ::: $a^3 - 3a^2b + 3ab^2 - b^3$ $a^3 + b^3$ factor karo ::: $(a+b)(a^2 - ab + b^2)$ $a^3 - b^3$ factor karo ::: $(a-b)(a^2 + ab + b^2)$ $(a+b)^3$ mein coefficients kya hain? ::: 1, 3, 3, 1 (Pascal's Triangle row 3) Sum of cubes ke liye SOAP mnemonic kya hai? ::: Binomial mein same sign, trinomial ke middle term mein opposite sign, aakhri term hamesha positive $(a+b)(a-b) = a^2 - b^2$ mein middle terms kyun gayab ho jaate hain? ::: Kyunki distributive property se expand karne par $+ab$ aur $-ab$ cancel ho jaate hain Sach ya Jhooth: $(a-b)^2 = a^2 - b^2$ :: Jhooth. $(a-b)^2 = a^2 - 2ab + b^2$ (middle term $-2ab$ zaroori hai) $x^2 - 49$ factor karo ::: $(x+7)(x-7)$, $a^2 - b^2 = (a+b)(a-b)$ use karke $27x^3 + 8$ factor karo ::: $(3x+2)(9x^2 - 6x + 4)$, sum of cubes use karke, jahan $27x^3 = (3x)^3$ aur $8 = 2^3$ $(x+5)^2 - (x-5)^2$ kya hai? :: $20x$ (difference of squares use karo ya dono expand karo) $a^3 - b^3$ ke trinomial factor mein middle term ka sign kya hai? ::: 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 ```