Algebraic Identities

Flashcards made from Maths Unplugged's crash course video. Suitable for those writing JEE. Answer only with definition (I've included cards for both ways around).

(a+b)²

a²+b²+2ab

(a-b)²

a²+b²-2ab

a²+b²+2ab

(a+b)²

a²+b²-2ab

(a-b)²

(a+b)(a-b)

a²-b²

a²-b²

(a+b)(a-b)

a^{m^n}

a^{m^n}

(a^m)^n

a^{mn}

(\Sigma a)²

\Sigma(a²)+2\Sigma(a_1a_2)

\Sigma(a²)+2\Sigma(a_1a_2)

(\Sigma a)²

\Sigma a_1a_2

\dfrac{(\Sigma a)²-\Sigma (a²)}{2} 

\dfrac{(\Sigma a)²-\Sigma (a²)}{2} 

\Sigma a_1a_2

\Sigma a_1a_2 = \dfrac{(\Sigma a)²-\Sigma (a²)}{2}

How is this derived?

By using identities like (a+b)² and (a+b+c)².

(1+a)(1+a²)(1+a^4)(1+a^8)…(1+a^n)

\dfrac{1-a^{2n}}{1-a}

\dfrac{1-a^{2n}}{1-a}

(1+a)(1+a²)(1+a^4)(1+a^8)…(1+a^n)

\dfrac{1-a^{2n}}{1-a}=(1+a)(1+a²)(1+a^4)(1+a^8)…(1+a^n)

How is this derived?

Multiple and divide RHS by (1-a), then you use (a+b)(a-b)=a²-b²) to cancel out all the consecutive terms.

(a+b+c)²

a²+b²+c²+2(ab+bc+ca)

a²+b²+c²+2(ab+bc+ca)

(a+b+c)²

(a+b+c+d)²

a²+b²+c²+d²+2(ab + ac + ad + bc + bd + cd)

(a+b)³

a³+b³+3ab(a+b)

a³+b³+3ab(a+b)

(a+b)³

a²+b²+c²+d²+2(ab + ac + ad + bc + bd + cd)

(a+b+c+d)²

a³+b³

give both forms of this.

(a+b)³-3ab(a+b)

(a+b)(a²+b²-ab)

(a+b)³-3ab(a+b)

a³+b³

(a+b)(a²+b²-ab)

a³+b³

(a-b)³

a³-b³-3ab(a-b)

a³-b³-3ab(a-b)

(a-b)³

a³-b³

give both forms

(a-b)³+3ab(a-b)

(a-b)(a²+b²+ab)

(a-b)(a²+b²+ab)

a³-b³

(a-b)³+3ab(a-b)

a³-b³

a²-ab+b²

Give all 3 forms

(a+b)²-3ab

(a-b)²+ab

\dfrac{a³+b³}{(a+b)}

\dfrac{a³+b³}{(a+b)}

a²-ab+b²

a²+ab+b²

give all 3 forms

(a+b)²-ab

(a-b)²+3ab

\dfrac{a³-b³}{a-b}

\dfrac{a³-b³}{a-b}

a²+ab+b²

a^4-b^4

(a²+b²)(a²-b²)

a^4+b^4

(a²+b²)²-2a²b²

a²+b²

(a+b)²-2ab

a^4+4b^4

(a²+2b²-2ab)(a²+2b²+2ab)

derive simply and simplify

a^4+a²+1

(a²+a+1)(a²-a+1)

(a²+a+1)(a²-a+1)

a^4+a²+1

a^8+a^4+1

(a^4+a²+1)(a^4-a²+1)

(a^4+a²+1)(a^4-a²+1)

a^8+a^4+1

ab+bc+ca

\Big(\dfrac1a+\dfrac1b+\dfrac1c\Big)abc

\Big(\dfrac1a+\dfrac1b+\dfrac1c\Big)abc

ab+bc+ca

a²+b²+c²

(a+b+c)²-2(ab+bc+ca)

(a+b+c)²-2(ab+bc+ca)

a²+b²+c²

a³+b³+c³

(a+b+c)(a²+b²+c²-ab-bc-ca) + 3abc

(a+b+c)(a²+b²+c²-ab-bc-ca) + 3abc

a³+b³+c³

If a+b+c=0, what is a³+b³+c³=?

a³+b³+c³=3abc

if  a³+b³+c³=3abc, then what can you infer? (2 things)

either a+b+c=0

or a=b=c

a²+b²+c²-ab-bc-ca

Give all 3 forms of this.

=\dfrac{1}{2}[(a-b)²+(b-c)²+(c-a)²]

=(a+b+c)²-3(ab+bc+ca)

=\dfrac{(a+b+c)³-3abc}{a+b+c}

if a²+b²+c²-ab-bc-ca=0, what can you infer (2 things)?

that a³+b³+c³=3abc

and (a-b)²+(b-c)²+(c-a)²=0 which means a=b=c

if a²+b²+c²=0, what can you infer?

a=b=c=0

(a²-b²)³+(b²-c²)³+(c²-a²)³=?

How do you derive this?

3\left(a^2-b^2\right)\left(b^2-c^2\right)\left(c^2-a^2\right)

This is like X³+Y³+Z³.

Since X+Y+Z=0, then X³+Y³+Z³=3XYZ.

What can you infer from (x-a)²+(y-b)²+(z-c)³=0?

x=a,y=b,z=c

What can you infer from (x-a)²+(x-b)²+(x-c)³=0?

a=b=c

if not, the equation cannot exist

abc+(ab+bc+ca)+(a+b+c)+1

(a+1)(b+1)(c+1)

(a+1)(b+1)(c+1)

abc+(ab+bc+ca)+(a+b+c)+1

abcd+(a+b+c+d)+(ab+ac+ad+bc+bd+cd)+(abc+abd+acd+bcd)+1

(a+1)(b+1)(c+1)(d+1)

(a_1+1)(a_2+1)(a_3+1)+…+(a_n+1)

1 + \Sigma a_1 +\Sigma (a_1a_2)+\Sigma (a_1a_2a_3)+…+(a_1a_2a_3…a_n)

1 + \Sigma a_1 +\Sigma (a_1a_2)+\Sigma (a_1a_2a_3)+…+(a_1a_2a_3…a_n)

(a_1+1)(a_2+1)(a_3+1)+…+(a_n+1)

(a+b+c)³

a³+b³+c³+3(a+b)(b+c)(c+a)

a³+b³+c³+3(a+b)(b+c)(c+a)

(a+b+c)³

(a+b+c)³=a³+b³+c³+3(a+b)(b+c)(c+a)

How do you derive this?

consider X=a+b and Y=c and solve (X+Y)³.

(a-b+c+d)(a+b-c+d)(a+b+c-d)(b+c+d-a)

How do you approach this?

(a+b+c+d-2b)(a+b+c+d-2c)(a+b+d+c-2d)(a+b+c+d-2a)

Let a+b+c+d=x)

then,

=(x-2a)(x-2b)(x-2c)(x-2d).