Algebraic Identities

This note has been used to help create the Algebraic Manipulation Identities wiki


An identity is an equality that holds true regardless of the values chosen for its variables.

For example, the identity (x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2 is always true regardless of the values of xx and yy.


Since identities are true for all valid values of its variables, one side of the equality can be swapped for the other. For example, we can replace any instance of (x+y)2(x+y)^2 with x2+2xy+y2x^2 + 2xy + y^2 and vice versa because (x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2 is an identity.

Clever use of identities offers shortcuts to many problems by making the algebra easier to manipulate. Below are lists of some common algebraic identities.

These identities are product formulas that are basic examples of the binomial theorem. (x+y)2=x2+2xy+y2(xy)2=x22xy+y2(x+y)3=x3+3x2y+3xy2+y3(xy)3=x33x2y+3xy2y3(x+y)4=x4+4x3y+6x2y2+4xy3+y4(xy)4=x44x3y+6x2y24xy3+y4\begin{aligned} (x+y)^2 &= x^2 + 2xy + y^2 \\ (x-y)^2 &= x^2 - 2xy + y^2 \\ (x+y)^3 &= x^3+3x^2y + 3xy^2 + y^3 \\ (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 \\ (x-y)^4 &= x^4 - 4x^3y + 6x^2y^2-4xy^3 + y^4 \end{aligned}

These identities are factoring formulas. Their more general forms are listed on the factorization page.

x2y2=(x+y)(xy)x3y3=(xy)(x2+xy+y2)x3+y3=(x+y)(x2xy+y2)x4y4=(x2y2)(x2+y2)\begin{aligned} x^2 - y^2 &= (x+y)(x-y) \\ x^3 - y^3 &= (x-y)(x^2+xy+y^2) \\ x^3 + y^3 &= (x+y)(x^2-xy+y^2) \\ x^4 - y^4 &= (x^2-y^2)(x^2+y^2) \end{aligned}

Application and Extensions

The identity 4(x+7)(2x1)=Ax2+Bx+C4(x+7)(2x-1)=Ax^2+Bx+C holds for all real values of xx. What is A+B+CA+B+C?

Multiplying out the left side of the identity we get 4(x+7)(2x1)=8x2+52x28.4(x+7)(2x-1)=8x^2+52x-28.
This expression must be equal to the right hand side of the identity, therefore
so A=8A=8, B=52B=52, and C=28C=-28, which gives us A+B+C=8+5228=32A+B+C=8+52-28=32. _\square


If A+B=8A+B=8 and AB=13AB=13, what is A3+B3A^3+B^3?

While you could solve for AA and BB, a more elegant solution exploits the identity (x+y)3=x3+3x2y+3xy2+y3(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
which can be rewritten as x3+y3=(x+y)33xy(x+y).x^3+y^3 = (x+y)^3-3xy(x+y).
Substituting in AA and BB for xx and yy we get A3+B3=(A+B)33AB(A+B)=(8)33(13)(8)=512312=200\begin{aligned} A^3+B^3 &= (A+B)^3-3AB(A+B) \\ &= (8)^3-3(13)(8) \\ &= 512-312 \\ &= 200 \quad_\square \end{aligned}

Note by Arron Kau
7 years, 4 months ago

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Somes part of the text bug it operations i cant see it?

Magus-Astra Equilibrium - 7 years, 3 months ago

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can anyone give me some notes of lessons lines and angles and triangles with inequalities and with sums

ṠḀḧḭtḧ Ṙṏẏal - 6 years, 11 months ago

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good explanation

Syed Hamza Khalid - 4 years, 4 months ago

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IF X>1, find the minimum value of 2log10x – logx0.01

Suraj Pandey - 7 years ago

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abdelrahman turky - 7 years ago

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your expaination is good

Francis Cabanting - 7 years ago

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Why minus to minus is not plus

Harsh Lohia - 6 years, 10 months ago

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